Automata and Group Theory - UPC Universitat Politècnica de...
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Automata and Group Theory
Enric VenturaDepartament de Matemàtica Aplicada III
Universitat Politècnica de Catalunya
AutoMatha ABCD Workshop, Bratislava,
November 25, 2008
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 1 / 69
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Outline
1 The friendly and unfriendly free group
2 The bijection between subgroups and automata
3 Several algorithmic applications
4 Algebraic extensions and Takahasi’s theorem
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 2 / 69
![Page 3: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/3.jpg)
Outline
1 The friendly and unfriendly free group
2 The bijection between subgroups and automata
3 Several algorithmic applications
4 Algebraic extensions and Takahasi’s theorem
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 3 / 69
![Page 4: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/4.jpg)
Definitions and notation
A = {a1, . . . , an} is a finite alphabet (n letters).A±1 = A ∪ A−1 = {a1, a−1
1 , . . . , an, a−1n }.
Usually, A = {a, b, c}.(A±1)∗ the free monoid on A±1 (words on A±1).1 denotes the empty word, and we have the notion of length.∼ is the eq. rel. generated by aia−1
i ∼ a−1i ai ∼ 1.
FA = (A±1)∗/ ∼ is the free group on A (words on A±1 modulo ∼).Every w ∈ A∗ has a unique reduced form, denoted w , (clearly w = w inFA, and w is the shortest word with this property). We also say w is areduced word.Again 1 denotes the empty word, and | · | the (shortest) length in FA:|1| = 0, |aba−1| = |abbb−1a−1| = 3, |uv | 6 |u|+ |v |.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 4 / 69
![Page 5: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/5.jpg)
Definitions and notation
A = {a1, . . . , an} is a finite alphabet (n letters).A±1 = A ∪ A−1 = {a1, a−1
1 , . . . , an, a−1n }.
Usually, A = {a, b, c}.(A±1)∗ the free monoid on A±1 (words on A±1).1 denotes the empty word, and we have the notion of length.∼ is the eq. rel. generated by aia−1
i ∼ a−1i ai ∼ 1.
FA = (A±1)∗/ ∼ is the free group on A (words on A±1 modulo ∼).Every w ∈ A∗ has a unique reduced form, denoted w , (clearly w = w inFA, and w is the shortest word with this property). We also say w is areduced word.Again 1 denotes the empty word, and | · | the (shortest) length in FA:|1| = 0, |aba−1| = |abbb−1a−1| = 3, |uv | 6 |u|+ |v |.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 4 / 69
![Page 6: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/6.jpg)
Definitions and notation
A = {a1, . . . , an} is a finite alphabet (n letters).A±1 = A ∪ A−1 = {a1, a−1
1 , . . . , an, a−1n }.
Usually, A = {a, b, c}.(A±1)∗ the free monoid on A±1 (words on A±1).1 denotes the empty word, and we have the notion of length.∼ is the eq. rel. generated by aia−1
i ∼ a−1i ai ∼ 1.
FA = (A±1)∗/ ∼ is the free group on A (words on A±1 modulo ∼).Every w ∈ A∗ has a unique reduced form, denoted w , (clearly w = w inFA, and w is the shortest word with this property). We also say w is areduced word.Again 1 denotes the empty word, and | · | the (shortest) length in FA:|1| = 0, |aba−1| = |abbb−1a−1| = 3, |uv | 6 |u|+ |v |.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 4 / 69
![Page 7: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/7.jpg)
Definitions and notation
A = {a1, . . . , an} is a finite alphabet (n letters).A±1 = A ∪ A−1 = {a1, a−1
1 , . . . , an, a−1n }.
Usually, A = {a, b, c}.(A±1)∗ the free monoid on A±1 (words on A±1).1 denotes the empty word, and we have the notion of length.∼ is the eq. rel. generated by aia−1
i ∼ a−1i ai ∼ 1.
FA = (A±1)∗/ ∼ is the free group on A (words on A±1 modulo ∼).Every w ∈ A∗ has a unique reduced form, denoted w , (clearly w = w inFA, and w is the shortest word with this property). We also say w is areduced word.Again 1 denotes the empty word, and | · | the (shortest) length in FA:|1| = 0, |aba−1| = |abbb−1a−1| = 3, |uv | 6 |u|+ |v |.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 4 / 69
![Page 8: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/8.jpg)
Definitions and notation
A = {a1, . . . , an} is a finite alphabet (n letters).A±1 = A ∪ A−1 = {a1, a−1
1 , . . . , an, a−1n }.
Usually, A = {a, b, c}.(A±1)∗ the free monoid on A±1 (words on A±1).1 denotes the empty word, and we have the notion of length.∼ is the eq. rel. generated by aia−1
i ∼ a−1i ai ∼ 1.
FA = (A±1)∗/ ∼ is the free group on A (words on A±1 modulo ∼).Every w ∈ A∗ has a unique reduced form, denoted w , (clearly w = w inFA, and w is the shortest word with this property). We also say w is areduced word.Again 1 denotes the empty word, and | · | the (shortest) length in FA:|1| = 0, |aba−1| = |abbb−1a−1| = 3, |uv | 6 |u|+ |v |.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 4 / 69
![Page 9: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/9.jpg)
Definitions and notation
A = {a1, . . . , an} is a finite alphabet (n letters).A±1 = A ∪ A−1 = {a1, a−1
1 , . . . , an, a−1n }.
Usually, A = {a, b, c}.(A±1)∗ the free monoid on A±1 (words on A±1).1 denotes the empty word, and we have the notion of length.∼ is the eq. rel. generated by aia−1
i ∼ a−1i ai ∼ 1.
FA = (A±1)∗/ ∼ is the free group on A (words on A±1 modulo ∼).Every w ∈ A∗ has a unique reduced form, denoted w , (clearly w = w inFA, and w is the shortest word with this property). We also say w is areduced word.Again 1 denotes the empty word, and | · | the (shortest) length in FA:|1| = 0, |aba−1| = |abbb−1a−1| = 3, |uv | 6 |u|+ |v |.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 4 / 69
![Page 10: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/10.jpg)
Definitions and notation
A = {a1, . . . , an} is a finite alphabet (n letters).A±1 = A ∪ A−1 = {a1, a−1
1 , . . . , an, a−1n }.
Usually, A = {a, b, c}.(A±1)∗ the free monoid on A±1 (words on A±1).1 denotes the empty word, and we have the notion of length.∼ is the eq. rel. generated by aia−1
i ∼ a−1i ai ∼ 1.
FA = (A±1)∗/ ∼ is the free group on A (words on A±1 modulo ∼).Every w ∈ A∗ has a unique reduced form, denoted w , (clearly w = w inFA, and w is the shortest word with this property). We also say w is areduced word.Again 1 denotes the empty word, and | · | the (shortest) length in FA:|1| = 0, |aba−1| = |abbb−1a−1| = 3, |uv | 6 |u|+ |v |.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 4 / 69
![Page 11: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/11.jpg)
Definitions and notation
A = {a1, . . . , an} is a finite alphabet (n letters).A±1 = A ∪ A−1 = {a1, a−1
1 , . . . , an, a−1n }.
Usually, A = {a, b, c}.(A±1)∗ the free monoid on A±1 (words on A±1).1 denotes the empty word, and we have the notion of length.∼ is the eq. rel. generated by aia−1
i ∼ a−1i ai ∼ 1.
FA = (A±1)∗/ ∼ is the free group on A (words on A±1 modulo ∼).Every w ∈ A∗ has a unique reduced form, denoted w , (clearly w = w inFA, and w is the shortest word with this property). We also say w is areduced word.Again 1 denotes the empty word, and | · | the (shortest) length in FA:|1| = 0, |aba−1| = |abbb−1a−1| = 3, |uv | 6 |u|+ |v |.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 4 / 69
![Page 12: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/12.jpg)
Definitions and notation
A = {a1, . . . , an} is a finite alphabet (n letters).A±1 = A ∪ A−1 = {a1, a−1
1 , . . . , an, a−1n }.
Usually, A = {a, b, c}.(A±1)∗ the free monoid on A±1 (words on A±1).1 denotes the empty word, and we have the notion of length.∼ is the eq. rel. generated by aia−1
i ∼ a−1i ai ∼ 1.
FA = (A±1)∗/ ∼ is the free group on A (words on A±1 modulo ∼).Every w ∈ A∗ has a unique reduced form, denoted w , (clearly w = w inFA, and w is the shortest word with this property). We also say w is areduced word.Again 1 denotes the empty word, and | · | the (shortest) length in FA:|1| = 0, |aba−1| = |abbb−1a−1| = 3, |uv | 6 |u|+ |v |.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 4 / 69
![Page 13: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/13.jpg)
The universal property
The universal property: given a group G and a mapping ϕ : A→ G, thereexists a unique group homomorphism Φ: FA → G such that the diagram
Aϕ //
ι
��
G
FA
∃!Φ
>>~~
~~
commutes (where ι is the inclusion map).Every group is a quotient of a free group
G = 〈a1, . . . , an | r1, . . . , rm〉 = FA/� r1, . . . , rm � .
So, the lattice of (normal) subgroups of FA is very important.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 5 / 69
![Page 14: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/14.jpg)
The universal property
The universal property: given a group G and a mapping ϕ : A→ G, thereexists a unique group homomorphism Φ: FA → G such that the diagram
Aϕ //
ι
��
G
FA
∃!Φ
>>~~
~~
commutes (where ι is the inclusion map).Every group is a quotient of a free group
G = 〈a1, . . . , an | r1, . . . , rm〉 = FA/� r1, . . . , rm � .
So, the lattice of (normal) subgroups of FA is very important.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 5 / 69
![Page 15: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/15.jpg)
The universal property
The universal property: given a group G and a mapping ϕ : A→ G, thereexists a unique group homomorphism Φ: FA → G such that the diagram
Aϕ //
ι
��
G
FA
∃!Φ
>>~~
~~
commutes (where ι is the inclusion map).Every group is a quotient of a free group
G = 〈a1, . . . , an | r1, . . . , rm〉 = FA/� r1, . . . , rm � .
So, the lattice of (normal) subgroups of FA is very important.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 5 / 69
![Page 16: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/16.jpg)
Comparison with linear algebra
vector spaces free groups
• K n f.d. K -vector space • Fn f.g. free group
• Every f.d. K -vectorspace is like this,
• Every group G is a quotientof a free group,
• K n ' K m ⇔ n = m, • Fn ' Fm ⇔ n = m,
• – • (Nielsen-Schreier) Every subgroupof a free group is free,
• Steinitz Lemma, • Not true,
• F 6 E ⇒ dim F 6 dim E , • Very false: Fℵ0 6 F2.
• A basis • The A-Stallings automata
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 6 / 69
![Page 17: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/17.jpg)
Comparison with linear algebra
vector spaces free groups
• K n f.d. K -vector space • Fn f.g. free group
• Every f.d. K -vectorspace is like this,
• Every group G is a quotientof a free group,
• K n ' K m ⇔ n = m, • Fn ' Fm ⇔ n = m,
• – • (Nielsen-Schreier) Every subgroupof a free group is free,
• Steinitz Lemma, • Not true,
• F 6 E ⇒ dim F 6 dim E , • Very false: Fℵ0 6 F2.
• A basis • The A-Stallings automata
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 6 / 69
![Page 18: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/18.jpg)
Comparison with linear algebra
vector spaces free groups
• K n f.d. K -vector space • Fn f.g. free group
• Every f.d. K -vectorspace is like this,
• Every group G is a quotientof a free group,
• K n ' K m ⇔ n = m, • Fn ' Fm ⇔ n = m,
• – • (Nielsen-Schreier) Every subgroupof a free group is free,
• Steinitz Lemma, • Not true,
• F 6 E ⇒ dim F 6 dim E , • Very false: Fℵ0 6 F2.
• A basis • The A-Stallings automata
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 6 / 69
![Page 19: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/19.jpg)
Comparison with linear algebra
vector spaces free groups
• K n f.d. K -vector space • Fn f.g. free group
• Every f.d. K -vectorspace is like this,
• Every group G is a quotientof a free group,
• K n ' K m ⇔ n = m, • Fn ' Fm ⇔ n = m,
• – • (Nielsen-Schreier) Every subgroupof a free group is free,
• Steinitz Lemma, • Not true,
• F 6 E ⇒ dim F 6 dim E , • Very false: Fℵ0 6 F2.
• A basis • The A-Stallings automata
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 6 / 69
![Page 20: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/20.jpg)
Comparison with linear algebra
vector spaces free groups
• K n f.d. K -vector space • Fn f.g. free group
• Every f.d. K -vectorspace is like this,
• Every group G is a quotientof a free group,
• K n ' K m ⇔ n = m, • Fn ' Fm ⇔ n = m,
• – • (Nielsen-Schreier) Every subgroupof a free group is free,
• Steinitz Lemma, • Not true,
• F 6 E ⇒ dim F 6 dim E , • Very false: Fℵ0 6 F2.
• A basis • The A-Stallings automata
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 6 / 69
![Page 21: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/21.jpg)
Comparison with linear algebra
vector spaces free groups
• K n f.d. K -vector space • Fn f.g. free group
• Every f.d. K -vectorspace is like this,
• Every group G is a quotientof a free group,
• K n ' K m ⇔ n = m, • Fn ' Fm ⇔ n = m,
• – • (Nielsen-Schreier) Every subgroupof a free group is free,
• Steinitz Lemma, • Not true,
• F 6 E ⇒ dim F 6 dim E , • Very false: Fℵ0 6 F2.
• A basis • The A-Stallings automata
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 6 / 69
![Page 22: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/22.jpg)
Comparison with linear algebra
vector spaces free groups
• K n f.d. K -vector space • Fn f.g. free group
• Every f.d. K -vectorspace is like this,
• Every group G is a quotientof a free group,
• K n ' K m ⇔ n = m, • Fn ' Fm ⇔ n = m,
• – • (Nielsen-Schreier) Every subgroupof a free group is free,
• Steinitz Lemma, • Not true,
• F 6 E ⇒ dim F 6 dim E , • Very false: Fℵ0 6 F2.
• A basis • The A-Stallings automata
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 6 / 69
![Page 23: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/23.jpg)
Comparison with linear algebra
vector spaces free groups
• K n f.d. K -vector space • Fn f.g. free group
• Every f.d. K -vectorspace is like this,
• Every group G is a quotientof a free group,
• K n ' K m ⇔ n = m, • Fn ' Fm ⇔ n = m,
• – • (Nielsen-Schreier) Every subgroupof a free group is free,
• Steinitz Lemma, • Not true,
• F 6 E ⇒ dim F 6 dim E , • Very false: Fℵ0 6 F2.
• A basis • The A-Stallings automata
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 6 / 69
![Page 24: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/24.jpg)
Comparison with linear algebra
vector spaces free groups
• K n f.d. K -vector space • Fn f.g. free group
• Every f.d. K -vectorspace is like this,
• Every group G is a quotientof a free group,
• K n ' K m ⇔ n = m, • Fn ' Fm ⇔ n = m,
• – • (Nielsen-Schreier) Every subgroupof a free group is free,
• Steinitz Lemma, • Not true,
• F 6 E ⇒ dim F 6 dim E , • Very false: Fℵ0 6 F2.
• A basis • The A-Stallings automata
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 6 / 69
![Page 25: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/25.jpg)
Comparison with linear algebra
vector spaces free groups
• K n f.d. K -vector space • Fn f.g. free group
• Every f.d. K -vectorspace is like this,
• Every group G is a quotientof a free group,
• K n ' K m ⇔ n = m, • Fn ' Fm ⇔ n = m,
• – • (Nielsen-Schreier) Every subgroupof a free group is free,
• Steinitz Lemma, • Not true,
• F 6 E ⇒ dim F 6 dim E , • Very false: Fℵ0 6 F2.
• A basis • The A-Stallings automata
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 6 / 69
![Page 26: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/26.jpg)
Comparison with linear algebra
vector spaces free groups
• K n f.d. K -vector space • Fn f.g. free group
• Every f.d. K -vectorspace is like this,
• Every group G is a quotientof a free group,
• K n ' K m ⇔ n = m, • Fn ' Fm ⇔ n = m,
• – • (Nielsen-Schreier) Every subgroupof a free group is free,
• Steinitz Lemma, • Not true,
• F 6 E ⇒ dim F 6 dim E , • Very false: Fℵ0 6 F2.
• A basis • The A-Stallings automata
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 6 / 69
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Comparison with linear algebra
vector spaces free groups
• K n f.d. K -vector space • Fn f.g. free group
• Every f.d. K -vectorspace is like this,
• Every group G is a quotientof a free group,
• K n ' K m ⇔ n = m, • Fn ' Fm ⇔ n = m,
• – • (Nielsen-Schreier) Every subgroupof a free group is free,
• Steinitz Lemma, • Not true,
• F 6 E ⇒ dim F 6 dim E , • Very false: Fℵ0 6 F2.
• A basis • The A-Stallings automata
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 6 / 69
![Page 28: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/28.jpg)
Comparison with linear algebra
vector spaces free groups
• K n f.d. K -vector space • Fn f.g. free group
• Every f.d. K -vectorspace is like this,
• Every group G is a quotientof a free group,
• K n ' K m ⇔ n = m, • Fn ' Fm ⇔ n = m,
• – • (Nielsen-Schreier) Every subgroupof a free group is free,
• Steinitz Lemma, • Not true,
• F 6 E ⇒ dim F 6 dim E , • Very false: Fℵ0 6 F2.
• A basis • The A-Stallings automata
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 6 / 69
![Page 29: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/29.jpg)
Comparison with linear algebra
vector spaces free groups
• K n f.d. K -vector space • Fn f.g. free group
• Every f.d. K -vectorspace is like this,
• Every group G is a quotientof a free group,
• K n ' K m ⇔ n = m, • Fn ' Fm ⇔ n = m,
• – • (Nielsen-Schreier) Every subgroupof a free group is free,
• Steinitz Lemma, • Not true,
• F 6 E ⇒ dim F 6 dim E , • Very false: Fℵ0 6 F2.
• A basis • The A-Stallings automata
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 6 / 69
![Page 30: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/30.jpg)
Comparison with linear algebra
vector spaces free groups
• K n f.d. K -vector space • Fn f.g. free group
• Every f.d. K -vectorspace is like this,
• Every group G is a quotientof a free group,
• K n ' K m ⇔ n = m, • Fn ' Fm ⇔ n = m,
• – • (Nielsen-Schreier) Every subgroupof a free group is free,
• Steinitz Lemma, • Not true,
• F 6 E ⇒ dim F 6 dim E , • Very false: Fℵ0 6 F2.
• A basis • The A-Stallings automata
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 6 / 69
![Page 31: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/31.jpg)
Outline
1 The friendly and unfriendly free group
2 The bijection between subgroups and automata
3 Several algorithmic applications
4 Algebraic extensions and Takahasi’s theorem
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 7 / 69
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Stallings automata
DefinitionA Stallings automata is a finite A-labeled oriented graph with a distinguishedvertex, (X , v), such that:
1- X is connected,2- no vertex of degree 1 except possibly v (X is a core-graph),3- no two edges with the same label go out of (or in to) the same vertex.
NO : •
a
��
b
���������������
• c // •a
** •
b
XX0000000000000
c
jj
YES : •
a
��
b
���������������
•a
** •
b
XX0000000000000
c
jj
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 8 / 69
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Stallings automata
DefinitionA Stallings automata is a finite A-labeled oriented graph with a distinguishedvertex, (X , v), such that:
1- X is connected,2- no vertex of degree 1 except possibly v (X is a core-graph),3- no two edges with the same label go out of (or in to) the same vertex.
NO : •
a
��
b
���������������
• c // •a
** •
b
XX0000000000000
c
jj
YES : •
a
��
b
���������������
•a
** •
b
XX0000000000000
c
jj
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 8 / 69
![Page 34: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/34.jpg)
Stallings automata
DefinitionA Stallings automata is a finite A-labeled oriented graph with a distinguishedvertex, (X , v), such that:
1- X is connected,2- no vertex of degree 1 except possibly v (X is a core-graph),3- no two edges with the same label go out of (or in to) the same vertex.
NO : •
a
��
b
���������������
• c // •a
** •
b
XX0000000000000
c
jj
YES : •
a
��
b
���������������
•a
** •
b
XX0000000000000
c
jj
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 8 / 69
![Page 35: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/35.jpg)
Stallings automata
In the influent paper
J. R. Stallings, Topology of finite graphs, Inventiones Math. 71 (1983),551-565,
Stallings (building on previous works) gave a bijection between finitelygenerated subgroups of FA and Stallings automata:
{f.g. subgroups of FA} ←→ {Stallings automata},
which is crucial for the modern understanding of the lattice of subgroups of FA.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 9 / 69
![Page 36: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/36.jpg)
Stallings automata
In the influent paper
J. R. Stallings, Topology of finite graphs, Inventiones Math. 71 (1983),551-565,
Stallings (building on previous works) gave a bijection between finitelygenerated subgroups of FA and Stallings automata:
{f.g. subgroups of FA} ←→ {Stallings automata},
which is crucial for the modern understanding of the lattice of subgroups of FA.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 9 / 69
![Page 37: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/37.jpg)
Stallings automata
In the influent paper
J. R. Stallings, Topology of finite graphs, Inventiones Math. 71 (1983),551-565,
Stallings (building on previous works) gave a bijection between finitelygenerated subgroups of FA and Stallings automata:
{f.g. subgroups of FA} ←→ {Stallings automata},
which is crucial for the modern understanding of the lattice of subgroups of FA.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 9 / 69
![Page 38: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/38.jpg)
Reading the subgroup from the automata
DefinitionTo any given (Stallings) automaton (X , v), we associate its fundamentalgroup:
π(X , v) = { labels of closed paths at v} 6 FA,
clearly, a subgroup of FA.
•
a
��
X= b
���������������
•a
** •
b
XX0000000000000
c
jj
π(X , •) = {1, a, a−1, bab, bc−1b,babab−1cb−1, . . .}
π(X , •) 63 bc−1bcaa
Membership problem in π(X , •) is solvable.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 10 / 69
![Page 39: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/39.jpg)
Reading the subgroup from the automata
DefinitionTo any given (Stallings) automaton (X , v), we associate its fundamentalgroup:
π(X , v) = { labels of closed paths at v} 6 FA,
clearly, a subgroup of FA.
•
a
��
X= b
���������������
•a
** •
b
XX0000000000000
c
jj
π(X , •) = {1, a, a−1, bab, bc−1b,babab−1cb−1, . . .}
π(X , •) 63 bc−1bcaa
Membership problem in π(X , •) is solvable.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 10 / 69
![Page 40: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/40.jpg)
Reading the subgroup from the automata
DefinitionTo any given (Stallings) automaton (X , v), we associate its fundamentalgroup:
π(X , v) = { labels of closed paths at v} 6 FA,
clearly, a subgroup of FA.
•
a
��
X= b
���������������
•a
** •
b
XX0000000000000
c
jj
π(X , •) = {1, a, a−1, bab, bc−1b,babab−1cb−1, . . .}
π(X , •) 63 bc−1bcaa
Membership problem in π(X , •) is solvable.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 10 / 69
![Page 41: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/41.jpg)
Reading the subgroup from the automata
DefinitionTo any given (Stallings) automaton (X , v), we associate its fundamentalgroup:
π(X , v) = { labels of closed paths at v} 6 FA,
clearly, a subgroup of FA.
•
a
��
X= b
���������������
•a
** •
b
XX0000000000000
c
jj
π(X , •) = {1, a, a−1, bab, bc−1b,babab−1cb−1, . . .}
π(X , •) 63 bc−1bcaa
Membership problem in π(X , •) is solvable.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 10 / 69
![Page 42: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/42.jpg)
Reading the subgroup from the automata
DefinitionTo any given (Stallings) automaton (X , v), we associate its fundamentalgroup:
π(X , v) = { labels of closed paths at v} 6 FA,
clearly, a subgroup of FA.
•
a
��
X= b
���������������
•a
** •
b
XX0000000000000
c
jj
π(X , •) = {1, a, a−1, bab, bc−1b,babab−1cb−1, . . .}
π(X , •) 63 bc−1bcaa
Membership problem in π(X , •) is solvable.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 10 / 69
![Page 43: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/43.jpg)
Reading the subgroup from the automata
DefinitionTo any given (Stallings) automaton (X , v), we associate its fundamentalgroup:
π(X , v) = { labels of closed paths at v} 6 FA,
clearly, a subgroup of FA.
•
a
��
X= b
���������������
•a
** •
b
XX0000000000000
c
jj
π(X , •) = {1, a, a−1, bab, bc−1b,babab−1cb−1, . . .}
π(X , •) 63 bc−1bcaa
Membership problem in π(X , •) is solvable.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 10 / 69
![Page 44: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/44.jpg)
Reading the subgroup from the automata
DefinitionTo any given (Stallings) automaton (X , v), we associate its fundamentalgroup:
π(X , v) = { labels of closed paths at v} 6 FA,
clearly, a subgroup of FA.
•
a
��
X= b
���������������
•a
** •
b
XX0000000000000
c
jj
π(X , •) = {1, a, a−1, bab, bc−1b,babab−1cb−1, . . .}
π(X , •) 63 bc−1bcaa
Membership problem in π(X , •) is solvable.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 10 / 69
![Page 45: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/45.jpg)
Reading the subgroup from the automata
DefinitionTo any given (Stallings) automaton (X , v), we associate its fundamentalgroup:
π(X , v) = { labels of closed paths at v} 6 FA,
clearly, a subgroup of FA.
•
a
��
X= b
���������������
•a
** •
b
XX0000000000000
c
jj
π(X , •) = {1, a, a−1, bab, bc−1b,babab−1cb−1, . . .}
π(X , •) 63 bc−1bcaa
Membership problem in π(X , •) is solvable.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 10 / 69
![Page 46: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/46.jpg)
Reading the subgroup from the automata
DefinitionTo any given (Stallings) automaton (X , v), we associate its fundamentalgroup:
π(X , v) = { labels of closed paths at v} 6 FA,
clearly, a subgroup of FA.
•
a
��
X= b
���������������
•a
** •
b
XX0000000000000
c
jj
π(X , •) = {1, a, a−1, bab, bc−1b,babab−1cb−1, . . .}
π(X , •) 63 bc−1bcaa
Membership problem in π(X , •) is solvable.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 10 / 69
![Page 47: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/47.jpg)
Reading the subgroup from the automata
DefinitionTo any given (Stallings) automaton (X , v), we associate its fundamentalgroup:
π(X , v) = { labels of closed paths at v} 6 FA,
clearly, a subgroup of FA.
•
a
��
X= b
���������������
•a
** •
b
XX0000000000000
c
jj
π(X , •) = {1, a, a−1, bab, bc−1b,babab−1cb−1, . . .}
π(X , •) 63 bc−1bcaa
Membership problem in π(X , •) is solvable.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 10 / 69
![Page 48: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/48.jpg)
A basis for π(X , v)
PropositionFor every Stallings automaton (X , v), the group π(X , v) is free of rankrk(π(X , v)) = 1− |VX |+ |EX |.
Proof:Take a maximal tree T in X .Write T [p, q] for the geodesic (i.e. the unique reduced path) in T from pto q.For every e ∈ EX − ET , xe = label(T [v , ιe] · e · T [τe, v ]) belongs toπ(X , v).Not difficult to see that {xe | e ∈ EX − ET} is a basis for π(X , v).And, |EX − ET | = |EX | − |ET |
= |EX | − (|VT | − 1) = 1− |VX |+ |EX |. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 11 / 69
![Page 49: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/49.jpg)
A basis for π(X , v)
PropositionFor every Stallings automaton (X , v), the group π(X , v) is free of rankrk(π(X , v)) = 1− |VX |+ |EX |.
Proof:Take a maximal tree T in X .Write T [p, q] for the geodesic (i.e. the unique reduced path) in T from pto q.For every e ∈ EX − ET , xe = label(T [v , ιe] · e · T [τe, v ]) belongs toπ(X , v).Not difficult to see that {xe | e ∈ EX − ET} is a basis for π(X , v).And, |EX − ET | = |EX | − |ET |
= |EX | − (|VT | − 1) = 1− |VX |+ |EX |. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 11 / 69
![Page 50: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/50.jpg)
A basis for π(X , v)
PropositionFor every Stallings automaton (X , v), the group π(X , v) is free of rankrk(π(X , v)) = 1− |VX |+ |EX |.
Proof:Take a maximal tree T in X .Write T [p, q] for the geodesic (i.e. the unique reduced path) in T from pto q.For every e ∈ EX − ET , xe = label(T [v , ιe] · e · T [τe, v ]) belongs toπ(X , v).Not difficult to see that {xe | e ∈ EX − ET} is a basis for π(X , v).And, |EX − ET | = |EX | − |ET |
= |EX | − (|VT | − 1) = 1− |VX |+ |EX |. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 11 / 69
![Page 51: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/51.jpg)
A basis for π(X , v)
PropositionFor every Stallings automaton (X , v), the group π(X , v) is free of rankrk(π(X , v)) = 1− |VX |+ |EX |.
Proof:Take a maximal tree T in X .Write T [p, q] for the geodesic (i.e. the unique reduced path) in T from pto q.For every e ∈ EX − ET , xe = label(T [v , ιe] · e · T [τe, v ]) belongs toπ(X , v).Not difficult to see that {xe | e ∈ EX − ET} is a basis for π(X , v).And, |EX − ET | = |EX | − |ET |
= |EX | − (|VT | − 1) = 1− |VX |+ |EX |. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 11 / 69
![Page 52: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/52.jpg)
A basis for π(X , v)
PropositionFor every Stallings automaton (X , v), the group π(X , v) is free of rankrk(π(X , v)) = 1− |VX |+ |EX |.
Proof:Take a maximal tree T in X .Write T [p, q] for the geodesic (i.e. the unique reduced path) in T from pto q.For every e ∈ EX − ET , xe = label(T [v , ιe] · e · T [τe, v ]) belongs toπ(X , v).Not difficult to see that {xe | e ∈ EX − ET} is a basis for π(X , v).And, |EX − ET | = |EX | − |ET |
= |EX | − (|VT | − 1) = 1− |VX |+ |EX |. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 11 / 69
![Page 53: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/53.jpg)
A basis for π(X , v)
PropositionFor every Stallings automaton (X , v), the group π(X , v) is free of rankrk(π(X , v)) = 1− |VX |+ |EX |.
Proof:Take a maximal tree T in X .Write T [p, q] for the geodesic (i.e. the unique reduced path) in T from pto q.For every e ∈ EX − ET , xe = label(T [v , ιe] · e · T [τe, v ]) belongs toπ(X , v).Not difficult to see that {xe | e ∈ EX − ET} is a basis for π(X , v).And, |EX − ET | = |EX | − |ET |
= |EX | − (|VT | − 1) = 1− |VX |+ |EX |. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 11 / 69
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Example
•
a
��
b
���������
•a
** •
b
XX0000000
c
jj
H = 〈 〉
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 12 / 69
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Example
•
a
��
b
���������
•a
** •
b
XX0000000
c
jj
H = 〈a, 〉
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 13 / 69
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Example
•
a
��
b
���������
•a
** •
b
XX0000000
c
jj
H = 〈a, bab, 〉
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 14 / 69
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Example
•
a
��
b
���������
•a
** •
b
XX0000000
c
jj
H = 〈a, bab, b−1cb−1〉
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 15 / 69
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Example
•
a
��
b
���������
•a
** •
b
XX0000000
c
jj
H = 〈a, bab, b−1cb−1〉rk(H) = 1− 3 + 5 = 3.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 16 / 69
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Example-2
· · · b // • b //
a
��• b //
a
��• b //
a
��• b //
a
��• b //
a
��• b //
a
��• b //
a
��· · ·
Fℵ0 ' H = 〈. . . , b−2ab2, b−1ab, a, bab−1, b2ab−2, . . .〉 6 F2.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 17 / 69
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Constructing the automata from the subgroup
In any automaton containing the following situation, for x ∈ A±1,
• x //
x&&NNNNNNNNNNNNN u
v
we can fold and identify vertices u and v to obtain
• x // u = v .
This operation, (X , v) (X ′, v), is called a Stallings folding.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 18 / 69
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Constructing the automata from the subgroup
In any automaton containing the following situation, for x ∈ A±1,
• x //
x&&NNNNNNNNNNNNN u
v
we can fold and identify vertices u and v to obtain
• x // u = v .
This operation, (X , v) (X ′, v), is called a Stallings folding.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 18 / 69
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Constructing the automata from the subgroup
In any automaton containing the following situation, for x ∈ A±1,
• x //
x&&NNNNNNNNNNNNN u
v
we can fold and identify vertices u and v to obtain
• x // u = v .
This operation, (X , v) (X ′, v), is called a Stallings folding.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 18 / 69
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Constructing the automata from the subgroup
Lemma (Stallings)If (X , v) (X ′, v ′) is a Stallings folding then π(X , v) = π(X ′, v ′).
Given a f.g. subgroup H = 〈w1, . . . wm〉 6 FA (we assume wi are reducedwords), do the following:
1- Draw the flower automaton,2- Perform successive foldings until obtaining a Stallings automaton,
denoted Γ(H).
Well defined?Need to see that the output does not depend on the process...
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 19 / 69
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Constructing the automata from the subgroup
Lemma (Stallings)If (X , v) (X ′, v ′) is a Stallings folding then π(X , v) = π(X ′, v ′).
Given a f.g. subgroup H = 〈w1, . . . wm〉 6 FA (we assume wi are reducedwords), do the following:
1- Draw the flower automaton,2- Perform successive foldings until obtaining a Stallings automaton,
denoted Γ(H).
Well defined?Need to see that the output does not depend on the process...
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 19 / 69
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Constructing the automata from the subgroup
Lemma (Stallings)If (X , v) (X ′, v ′) is a Stallings folding then π(X , v) = π(X ′, v ′).
Given a f.g. subgroup H = 〈w1, . . . wm〉 6 FA (we assume wi are reducedwords), do the following:
1- Draw the flower automaton,2- Perform successive foldings until obtaining a Stallings automaton,
denoted Γ(H).
Well defined?Need to see that the output does not depend on the process...
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 19 / 69
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Constructing the automata from the subgroup
Lemma (Stallings)If (X , v) (X ′, v ′) is a Stallings folding then π(X , v) = π(X ′, v ′).
Given a f.g. subgroup H = 〈w1, . . . wm〉 6 FA (we assume wi are reducedwords), do the following:
1- Draw the flower automaton,2- Perform successive foldings until obtaining a Stallings automaton,
denoted Γ(H).
Well defined?Need to see that the output does not depend on the process...
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 19 / 69
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Constructing the automata from the subgroup
Lemma (Stallings)If (X , v) (X ′, v ′) is a Stallings folding then π(X , v) = π(X ′, v ′).
Given a f.g. subgroup H = 〈w1, . . . wm〉 6 FA (we assume wi are reducedwords), do the following:
1- Draw the flower automaton,2- Perform successive foldings until obtaining a Stallings automaton,
denoted Γ(H).
Well defined?Need to see that the output does not depend on the process...
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 19 / 69
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Example: H = 〈baba−1, aba−1, aba2〉
• a // •
b
��• a // •
b
OO
a //
a
��???
????
????
????
?
a
��
a
������
����
����
����
•
•
a
??����������������•
boo • •
boo
Flower(H)
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 20 / 69
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Example: H = 〈baba−1, aba−1, aba2〉
• a // •
b
��• a // •
b
OO
a//
a
��???
????
????
????
?
a
��
a
������
����
����
����
•
•
a
??����������������•
boo • •
boo
Flower(H)
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 21 / 69
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Example: H = 〈baba−1, aba−1, aba2〉
• a // •
b
��•
b
OO
a // •
b
��
bpp
•
a
OO
•a
oo
Folding #1
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 22 / 69
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Example: H = 〈baba−1, aba−1, aba2〉
• a // •
b
��•
b
OO
a // •
b
��
bpp
•
a
OO
•a
oo
Folding #1.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 23 / 69
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Example: H = 〈baba−1, aba−1, aba2〉
•
a
��???
????
????
????
?
•
b
OO
a // •
a
������
����
����
����
bpp
•
a
OO
Folding #2.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 24 / 69
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Example: H = 〈baba−1, aba−1, aba2〉
•
a
��???
????
????
????
?
•
b
OO
a// •
a
������
����
����
����
bpp
•
a
OO
Folding #2.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 25 / 69
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Example: H = 〈baba−1, aba−1, aba2〉
• a //b
.. •
a
������
����
����
����
bpp
•
a
OO
Folding #3. Γ(H)
By Stallings Lemma, π(Γ(H), •) = 〈baba−1, aba−1, aba2〉
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 26 / 69
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Example: H = 〈baba−1, aba−1, aba2〉
• a //b
.. •
a
������
����
����
����
bpp
•
a
OO
Folding #3. Γ(H)
By Stallings Lemma, π(Γ(H), •) = 〈baba−1, aba−1, aba2〉
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 26 / 69
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Example: H = 〈baba−1, aba−1, aba2〉
• a //b
.. •
a
������
����
����
����
bpp
•
a
OO
Folding #3. Γ(H)
By Stallings Lemma, π(Γ(H), •) = 〈baba−1, aba−1, aba2〉= 〈b, aba−1, a3〉
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 27 / 69
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Local confluence
PropositionThe automaton Γ(H) does not depend on the sequence of foldings
Proof:Suppose (X , v) (X ′, v ′) is a single folding of 2 edges
If p x //
x&&MMMMMMMMMMMMM q
r
in (X , v), then p′ x //
x&&MMMMMMMMMMMMM q′
r ′
in (X ′, v ′) (possibly
with q′ = r ′).So, we get local confluence:
(X , v)∀ //
∀��
π(X ′, v ′)
∃�����
π(X ′′, v ′′) ∃ //___ π(X ′′′, v ′′′)
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 28 / 69
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Local confluence
PropositionThe automaton Γ(H) does not depend on the sequence of foldings
Proof:Suppose (X , v) (X ′, v ′) is a single folding of 2 edges
If p x //
x&&MMMMMMMMMMMMM q
r
in (X , v), then p′ x //
x&&MMMMMMMMMMMMM q′
r ′
in (X ′, v ′) (possibly
with q′ = r ′).So, we get local confluence:
(X , v)∀ //
∀��
π(X ′, v ′)
∃�����
π(X ′′, v ′′) ∃ //___ π(X ′′′, v ′′′)
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 28 / 69
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Local confluence
PropositionThe automaton Γ(H) does not depend on the sequence of foldings
Proof:Suppose (X , v) (X ′, v ′) is a single folding of 2 edges
If p x //
x&&MMMMMMMMMMMMM q
r
in (X , v), then p′ x //
x&&MMMMMMMMMMMMM q′
r ′
in (X ′, v ′) (possibly
with q′ = r ′).So, we get local confluence:
(X , v)∀ //
∀��
π(X ′, v ′)
∃�����
π(X ′′, v ′′) ∃ //___ π(X ′′′, v ′′′)
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 28 / 69
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Local confluence
PropositionThe automaton Γ(H) does not depend on the sequence of foldings
Proof:Suppose (X , v) (X ′, v ′) is a single folding of 2 edges
If p x //
x&&MMMMMMMMMMMMM q
r
in (X , v), then p′ x //
x&&MMMMMMMMMMMMM q′
r ′
in (X ′, v ′) (possibly
with q′ = r ′).So, we get local confluence:
(X , v)∀ //
∀��
π(X ′, v ′)
∃�����
π(X ′′, v ′′) ∃ //___ π(X ′′′, v ′′′)
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 28 / 69
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The confluence grid
π(X00, v) //
��
π(X01, v) //
�����
. . . // π(X0p, v)
π(X10, v)
��
//___ π(X11, v)
...
��π(Xq0, v)
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 29 / 69
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The confluence grid
π(X00, v) //
��
π(X01, v) //
�����
. . . // π(X0p, v)
�����
π(X10, v)
��
//___ π(X11, v) //___ . . . //___ π(X1p, v)
...
��π(Xq0, v)
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 30 / 69
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The confluence grid
π(X00, v) //
��
π(X01, v) //
�����
. . . // π(X0p, v)
�����
π(X10, v)
��
//___ π(X11, v)
�����
//___ . . . //___ π(X1p, v)
�����
......
...
�� �����
�����
π(Xq0, v) //___ π(Xq1, v) //___ . . . //___ π(Xqp, v)
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 31 / 69
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Confluence
Hence, we have confluence:
(X , v)∀ +3
∀��
π(X ′, v ′)
∃�����
���
π(X ′′, v ′′) ∃ +3___ ___ π(X ′′′, v ′′′),
where⇒ stands for an arbitrary sequence of foldings.
Finally, edge-reducing + confluence implies unique output. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 32 / 69
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Confluence
Hence, we have confluence:
(X , v)∀ +3
∀��
π(X ′, v ′)
∃�����
���
π(X ′′, v ′′) ∃ +3___ ___ π(X ′′′, v ′′′),
where⇒ stands for an arbitrary sequence of foldings.
Finally, edge-reducing + confluence implies unique output. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 32 / 69
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Independence from the generators
PropositionThe automaton Γ(H) does not depend on the generators of H.
Proof:Suppose H = 〈w1, . . . , wp〉 = 〈w ′
1, . . . , w ′q〉 and let Γ(H) and Γ′(H) be the
Stallings automata obtained from each set of generators.Consider the double flower
•
w1
···
��
wp
��
w ′1
...
NN
w ′q
\\
whose fundamental group is 〈w1, . . . , wp, w ′1, . . . , w ′
q〉 = H.Now, fold in the two natural ways:
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 33 / 69
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Independence from the generators
PropositionThe automaton Γ(H) does not depend on the generators of H.
Proof:Suppose H = 〈w1, . . . , wp〉 = 〈w ′
1, . . . , w ′q〉 and let Γ(H) and Γ′(H) be the
Stallings automata obtained from each set of generators.Consider the double flower
•
w1
···
��
wp
��
w ′1
...
NN
w ′q
\\
whose fundamental group is 〈w1, . . . , wp, w ′1, . . . , w ′
q〉 = H.Now, fold in the two natural ways:
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 33 / 69
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Independence from the generators
PropositionThe automaton Γ(H) does not depend on the generators of H.
Proof:Suppose H = 〈w1, . . . , wp〉 = 〈w ′
1, . . . , w ′q〉 and let Γ(H) and Γ′(H) be the
Stallings automata obtained from each set of generators.Consider the double flower
•
w1
···
��
wp
��
w ′1
...
NN
w ′q
\\
whose fundamental group is 〈w1, . . . , wp, w ′1, . . . , w ′
q〉 = H.Now, fold in the two natural ways:
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 33 / 69
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Independence from the generators
PropositionThe automaton Γ(H) does not depend on the generators of H.
Proof:Suppose H = 〈w1, . . . , wp〉 = 〈w ′
1, . . . , w ′q〉 and let Γ(H) and Γ′(H) be the
Stallings automata obtained from each set of generators.Consider the double flower
•
w1
···
��
wp
��
w ′1
...
NN
w ′q
\\
whose fundamental group is 〈w1, . . . , wp, w ′1, . . . , w ′
q〉 = H.Now, fold in the two natural ways:
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 33 / 69
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Independence from the generators
Γ(H)•
w ′1
...
LL
w ′q
YY
•
w1
···
��
wp
��
w ′1
...
NN
w ′q
\\
19kkkkkkkkkkkkkkkkkkk
kkkkkkkkkkkkkkkkkkk
%-SSSSSSSSSSSSSSSSSS
SSSSSSSSSSSSSSSSSS
•Γ′(H)
w1
···��
wp
��
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 34 / 69
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Independence from the generators
Γ(H)•
w ′1
...
LL
w ′q
YY+3 Γ(H)
•
w1
···
��
wp
��
w ′1
...
NN
w ′q
\\
19kkkkkkkkkkkkkkkkkkk
kkkkkkkkkkkkkkkkkkk
%-SSSSSSSSSSSSSSSSSS
SSSSSSSSSSSSSSSSSS
•Γ′(H)
w1
···��
wp
��+3 Γ′(H)
Lemma (Useless-w)If H 6fg FA and w ∈ H then, attaching a petal labeled w to the basepoint ofΓ(H) and folding, we obtain again Γ(H).
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 35 / 69
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Independence from the generators
Γ(H)•
w ′1
...
LL
w ′q
YY+3 Γ(H)
•
w1
···
��
wp
��
w ′1
...
NN
w ′q
\\
19lllllllllllllllllll
lllllllllllllllllll
$,RRRRRRRRRRRRRRRRRR
RRRRRRRRRRRRRRRRRR ‖
•Γ′(H)
w1
···��
wp
��+3 Γ′(H)
Lemma (Useless-w)If H 6fg FA and w ∈ H then, attaching a petal labeled w to the basepoint ofΓ(H) and folding, we obtain again Γ(H).
�
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 36 / 69
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The bijection
TheoremThe following is a bijection:
{f.g. subgroups of FA} ←→ {Stallings automata}H → Γ(H)
π(X , v) ← (X , v)
Proof:By Stallings Lemma, it is clear that π(Γ(H)) = H.Let (X , v) be a Stallings automata, and π(X , v) = = 〈w1, . . . , wp〉.Let (Y , v) be the automata obtained by attaching petals labeledw1, . . . , wp to the vertex v of (X , v).By the useless-w Lemma, (Y , v) can be folded to both (X , v) andΓ(π(X , v)). And both are completely folded. Hence, Γ(π(X , v)) = (X , v).�
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 37 / 69
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The bijection
TheoremThe following is a bijection:
{f.g. subgroups of FA} ←→ {Stallings automata}H → Γ(H)
π(X , v) ← (X , v)
Proof:By Stallings Lemma, it is clear that π(Γ(H)) = H.Let (X , v) be a Stallings automata, and π(X , v) = = 〈w1, . . . , wp〉.Let (Y , v) be the automata obtained by attaching petals labeledw1, . . . , wp to the vertex v of (X , v).By the useless-w Lemma, (Y , v) can be folded to both (X , v) andΓ(π(X , v)). And both are completely folded. Hence, Γ(π(X , v)) = (X , v).�
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 37 / 69
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The bijection
TheoremThe following is a bijection:
{f.g. subgroups of FA} ←→ {Stallings automata}H → Γ(H)
π(X , v) ← (X , v)
Proof:By Stallings Lemma, it is clear that π(Γ(H)) = H.Let (X , v) be a Stallings automata, and π(X , v) = = 〈w1, . . . , wp〉.Let (Y , v) be the automata obtained by attaching petals labeledw1, . . . , wp to the vertex v of (X , v).By the useless-w Lemma, (Y , v) can be folded to both (X , v) andΓ(π(X , v)). And both are completely folded. Hence, Γ(π(X , v)) = (X , v).�
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 37 / 69
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The bijection
TheoremThe following is a bijection:
{f.g. subgroups of FA} ←→ {Stallings automata}H → Γ(H)
π(X , v) ← (X , v)
Proof:By Stallings Lemma, it is clear that π(Γ(H)) = H.Let (X , v) be a Stallings automata, and π(X , v) = = 〈w1, . . . , wp〉.Let (Y , v) be the automata obtained by attaching petals labeledw1, . . . , wp to the vertex v of (X , v).By the useless-w Lemma, (Y , v) can be folded to both (X , v) andΓ(π(X , v)). And both are completely folded. Hence, Γ(π(X , v)) = (X , v).�
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 37 / 69
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The bijection
TheoremThe following is a bijection:
{f.g. subgroups of FA} ←→ {Stallings automata}H → Γ(H)
π(X , v) ← (X , v)
Proof:By Stallings Lemma, it is clear that π(Γ(H)) = H.Let (X , v) be a Stallings automata, and π(X , v) = = 〈w1, . . . , wp〉.Let (Y , v) be the automata obtained by attaching petals labeledw1, . . . , wp to the vertex v of (X , v).By the useless-w Lemma, (Y , v) can be folded to both (X , v) andΓ(π(X , v)). And both are completely folded. Hence, Γ(π(X , v)) = (X , v).�
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 37 / 69
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Nielsen-Schreier Theorem
Corollary (Nielsen-Schreier)Every subgroup of FA is free.
We have proved the finitely generated case, but everything extends easilyto the general case.
The original proof (1920’s) is combinatorial and much more technical.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 38 / 69
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Nielsen-Schreier Theorem
Corollary (Nielsen-Schreier)Every subgroup of FA is free.
We have proved the finitely generated case, but everything extends easilyto the general case.
The original proof (1920’s) is combinatorial and much more technical.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 38 / 69
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Nielsen-Schreier Theorem
Corollary (Nielsen-Schreier)Every subgroup of FA is free.
We have proved the finitely generated case, but everything extends easilyto the general case.
The original proof (1920’s) is combinatorial and much more technical.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 38 / 69
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Outline
1 The friendly and unfriendly free group
2 The bijection between subgroups and automata
3 Several algorithmic applications
4 Algebraic extensions and Takahasi’s theorem
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 39 / 69
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Membership & containment
(Membership)Does w belong to H = 〈w1, . . . , wm〉 ?
Construct Γ(H),Check whether w is readable as a closed path in Γ(H) (at the basepoint).
(Containment)Given H = 〈w1, . . . , wm〉 and K = 〈v1, . . . , vn〉, is H 6 K ?
Construct Γ(K ),Check whether all the wi ’s are readable as closed paths in Γ(H) (at thebasepoint).
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 40 / 69
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Membership & containment
(Membership)Does w belong to H = 〈w1, . . . , wm〉 ?
Construct Γ(H),Check whether w is readable as a closed path in Γ(H) (at the basepoint).
(Containment)Given H = 〈w1, . . . , wm〉 and K = 〈v1, . . . , vn〉, is H 6 K ?
Construct Γ(K ),Check whether all the wi ’s are readable as closed paths in Γ(H) (at thebasepoint).
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 40 / 69
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Membership & containment
(Membership)Does w belong to H = 〈w1, . . . , wm〉 ?
Construct Γ(H),Check whether w is readable as a closed path in Γ(H) (at the basepoint).
(Containment)Given H = 〈w1, . . . , wm〉 and K = 〈v1, . . . , vn〉, is H 6 K ?
Construct Γ(K ),Check whether all the wi ’s are readable as closed paths in Γ(H) (at thebasepoint).
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 40 / 69
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Membership & containment
(Membership)Does w belong to H = 〈w1, . . . , wm〉 ?
Construct Γ(H),Check whether w is readable as a closed path in Γ(H) (at the basepoint).
(Containment)Given H = 〈w1, . . . , wm〉 and K = 〈v1, . . . , vn〉, is H 6 K ?
Construct Γ(K ),Check whether all the wi ’s are readable as closed paths in Γ(H) (at thebasepoint).
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 40 / 69
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Basis & conjugacy
(Computing a basis)Given H = 〈w1, . . . , wm〉, find a basis for H.
Construct Γ(H),Choose a maximal tree,Read the corresponding basis.
(Conjugacy)Given H = 〈w1, . . . , wm〉 and K = 〈v1, . . . , vn〉, are they conjugate (i.e. Hx = Kfor some x ∈ FA) ?
Construct Γ(H) and Γ(K ),Check whether the are “equal" up to the basepoint.Every path between the two basepoints spells a valid x .
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 41 / 69
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Basis & conjugacy
(Computing a basis)Given H = 〈w1, . . . , wm〉, find a basis for H.
Construct Γ(H),Choose a maximal tree,Read the corresponding basis.
(Conjugacy)Given H = 〈w1, . . . , wm〉 and K = 〈v1, . . . , vn〉, are they conjugate (i.e. Hx = Kfor some x ∈ FA) ?
Construct Γ(H) and Γ(K ),Check whether the are “equal" up to the basepoint.Every path between the two basepoints spells a valid x .
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 41 / 69
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Basis & conjugacy
(Computing a basis)Given H = 〈w1, . . . , wm〉, find a basis for H.
Construct Γ(H),Choose a maximal tree,Read the corresponding basis.
(Conjugacy)Given H = 〈w1, . . . , wm〉 and K = 〈v1, . . . , vn〉, are they conjugate (i.e. Hx = Kfor some x ∈ FA) ?
Construct Γ(H) and Γ(K ),Check whether the are “equal" up to the basepoint.Every path between the two basepoints spells a valid x .
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 41 / 69
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Basis & conjugacy
(Computing a basis)Given H = 〈w1, . . . , wm〉, find a basis for H.
Construct Γ(H),Choose a maximal tree,Read the corresponding basis.
(Conjugacy)Given H = 〈w1, . . . , wm〉 and K = 〈v1, . . . , vn〉, are they conjugate (i.e. Hx = Kfor some x ∈ FA) ?
Construct Γ(H) and Γ(K ),Check whether the are “equal" up to the basepoint.Every path between the two basepoints spells a valid x .
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 41 / 69
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Finite index subgroups
(Finite index)Given H = 〈w1, . . . , wm〉, is H 6f .i. FA ? If yes, find a set of cosetrepresentatives.
→ For u ∈ VΓ(H), choose p (the label of) a path from • to u; then,
{labels of paths from • to u} = π(Γ(H), •) · p = H · p
is a coset of FA/H.→ FA/H is in bijection with the set of vertices of the “extended Γ(H)”
Construct Γ(H),Check whether Γ(H) is complete (i.e. every letter going in and out ofevery vertex),Choose a maximal tree T in Γ(H),{T [•, v ] | v ∈ VΓ(H)} is a set of coset reps. for H 6f .i. FA.
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Finite index subgroups
(Finite index)Given H = 〈w1, . . . , wm〉, is H 6f .i. FA ? If yes, find a set of cosetrepresentatives.
→ For u ∈ VΓ(H), choose p (the label of) a path from • to u; then,
{labels of paths from • to u} = π(Γ(H), •) · p = H · p
is a coset of FA/H.→ FA/H is in bijection with the set of vertices of the “extended Γ(H)”
Construct Γ(H),Check whether Γ(H) is complete (i.e. every letter going in and out ofevery vertex),Choose a maximal tree T in Γ(H),{T [•, v ] | v ∈ VΓ(H)} is a set of coset reps. for H 6f .i. FA.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 42 / 69
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Finite index subgroups
(Finite index)Given H = 〈w1, . . . , wm〉, is H 6f .i. FA ? If yes, find a set of cosetrepresentatives.
→ For u ∈ VΓ(H), choose p (the label of) a path from • to u; then,
{labels of paths from • to u} = π(Γ(H), •) · p = H · p
is a coset of FA/H.→ FA/H is in bijection with the set of vertices of the “extended Γ(H)”
Construct Γ(H),Check whether Γ(H) is complete (i.e. every letter going in and out ofevery vertex),Choose a maximal tree T in Γ(H),{T [•, v ] | v ∈ VΓ(H)} is a set of coset reps. for H 6f .i. FA.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 42 / 69
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Finite index subgroups
(Finite index)Given H = 〈w1, . . . , wm〉, is H 6f .i. FA ? If yes, find a set of cosetrepresentatives.
→ For u ∈ VΓ(H), choose p (the label of) a path from • to u; then,
{labels of paths from • to u} = π(Γ(H), •) · p = H · p
is a coset of FA/H.→ FA/H is in bijection with the set of vertices of the “extended Γ(H)”
Construct Γ(H),Check whether Γ(H) is complete (i.e. every letter going in and out ofevery vertex),Choose a maximal tree T in Γ(H),{T [•, v ] | v ∈ VΓ(H)} is a set of coset reps. for H 6f .i. FA.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 42 / 69
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Example
H = 〈b, ac, c−1a, cac−1, c−1bc−1, cbc, c4, c2ac−2, c2bc−2〉
Γ(H) = •b ## a
**
c
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NN
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 43 / 69
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Example
H = 〈b, ac, c−1a, cac−1, c−1bc−1, cbc, c4, c2ac−2, c2bc−2〉
Γ(H) = •b ## a
**
c
��000
0000
0000
00•
a
jj
b
��
c
��•
a
�� bppcoo
•
b
@@
c
99tttttttttttttttttttttt
a
NN
F3 = H t Hc t Ha t Hac−1.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 44 / 69
![Page 116: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/116.jpg)
More on finite index
(Schreier index formula)If H 6f .i. FA is of index [F : H], then r(H) = 1 + [F : H] · (r(FA)− 1).
Proof:
r(H) = 1− |VΓ(H)|+ |EΓ(H)| = 1− |VΓ(H)|+ |A| · |VΓ(H)|= 1 + |VΓ(H)| · (|A| − 1) = 1 + [F : H] · (r(FA)− 1). �
Theorem (M. Hall)Every f.g. subgroup H 6fg FA is a free factor of a finite index one,H 6ff H ∗ L 6f .i. FA.
Proof:Compute Γ(H) from a generating set,Locate the “missing” heads and tails of edges (in equal number for everyletter),Add new edges until having a complete automata (Y , v),Clearly, H = π(Γ(H)) 6ff π(Y , v) 6f .i. FA. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 45 / 69
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More on finite index
(Schreier index formula)If H 6f .i. FA is of index [F : H], then r(H) = 1 + [F : H] · (r(FA)− 1).
Proof:
r(H) = 1− |VΓ(H)|+ |EΓ(H)| = 1− |VΓ(H)|+ |A| · |VΓ(H)|= 1 + |VΓ(H)| · (|A| − 1) = 1 + [F : H] · (r(FA)− 1). �
Theorem (M. Hall)Every f.g. subgroup H 6fg FA is a free factor of a finite index one,H 6ff H ∗ L 6f .i. FA.
Proof:Compute Γ(H) from a generating set,Locate the “missing” heads and tails of edges (in equal number for everyletter),Add new edges until having a complete automata (Y , v),Clearly, H = π(Γ(H)) 6ff π(Y , v) 6f .i. FA. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 45 / 69
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More on finite index
(Schreier index formula)If H 6f .i. FA is of index [F : H], then r(H) = 1 + [F : H] · (r(FA)− 1).
Proof:
r(H) = 1− |VΓ(H)|+ |EΓ(H)| = 1− |VΓ(H)|+ |A| · |VΓ(H)|= 1 + |VΓ(H)| · (|A| − 1) = 1 + [F : H] · (r(FA)− 1). �
Theorem (M. Hall)Every f.g. subgroup H 6fg FA is a free factor of a finite index one,H 6ff H ∗ L 6f .i. FA.
Proof:Compute Γ(H) from a generating set,Locate the “missing” heads and tails of edges (in equal number for everyletter),Add new edges until having a complete automata (Y , v),Clearly, H = π(Γ(H)) 6ff π(Y , v) 6f .i. FA. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 45 / 69
![Page 119: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/119.jpg)
More on finite index
(Schreier index formula)If H 6f .i. FA is of index [F : H], then r(H) = 1 + [F : H] · (r(FA)− 1).
Proof:
r(H) = 1− |VΓ(H)|+ |EΓ(H)| = 1− |VΓ(H)|+ |A| · |VΓ(H)|= 1 + |VΓ(H)| · (|A| − 1) = 1 + [F : H] · (r(FA)− 1). �
Theorem (M. Hall)Every f.g. subgroup H 6fg FA is a free factor of a finite index one,H 6ff H ∗ L 6f .i. FA.
Proof:Compute Γ(H) from a generating set,Locate the “missing” heads and tails of edges (in equal number for everyletter),Add new edges until having a complete automata (Y , v),Clearly, H = π(Γ(H)) 6ff π(Y , v) 6f .i. FA. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 45 / 69
![Page 120: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/120.jpg)
More on finite index
(Schreier index formula)If H 6f .i. FA is of index [F : H], then r(H) = 1 + [F : H] · (r(FA)− 1).
Proof:
r(H) = 1− |VΓ(H)|+ |EΓ(H)| = 1− |VΓ(H)|+ |A| · |VΓ(H)|= 1 + |VΓ(H)| · (|A| − 1) = 1 + [F : H] · (r(FA)− 1). �
Theorem (M. Hall)Every f.g. subgroup H 6fg FA is a free factor of a finite index one,H 6ff H ∗ L 6f .i. FA.
Proof:Compute Γ(H) from a generating set,Locate the “missing” heads and tails of edges (in equal number for everyletter),Add new edges until having a complete automata (Y , v),Clearly, H = π(Γ(H)) 6ff π(Y , v) 6f .i. FA. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 45 / 69
![Page 121: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/121.jpg)
More on finite index
(Schreier index formula)If H 6f .i. FA is of index [F : H], then r(H) = 1 + [F : H] · (r(FA)− 1).
Proof:
r(H) = 1− |VΓ(H)|+ |EΓ(H)| = 1− |VΓ(H)|+ |A| · |VΓ(H)|= 1 + |VΓ(H)| · (|A| − 1) = 1 + [F : H] · (r(FA)− 1). �
Theorem (M. Hall)Every f.g. subgroup H 6fg FA is a free factor of a finite index one,H 6ff H ∗ L 6f .i. FA.
Proof:Compute Γ(H) from a generating set,Locate the “missing” heads and tails of edges (in equal number for everyletter),Add new edges until having a complete automata (Y , v),Clearly, H = π(Γ(H)) 6ff π(Y , v) 6f .i. FA. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 45 / 69
![Page 122: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/122.jpg)
More on finite index
(Schreier index formula)If H 6f .i. FA is of index [F : H], then r(H) = 1 + [F : H] · (r(FA)− 1).
Proof:
r(H) = 1− |VΓ(H)|+ |EΓ(H)| = 1− |VΓ(H)|+ |A| · |VΓ(H)|= 1 + |VΓ(H)| · (|A| − 1) = 1 + [F : H] · (r(FA)− 1). �
Theorem (M. Hall)Every f.g. subgroup H 6fg FA is a free factor of a finite index one,H 6ff H ∗ L 6f .i. FA.
Proof:Compute Γ(H) from a generating set,Locate the “missing” heads and tails of edges (in equal number for everyletter),Add new edges until having a complete automata (Y , v),Clearly, H = π(Γ(H)) 6ff π(Y , v) 6f .i. FA. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 45 / 69
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Example
H = 〈b, cbc, c2bc−2〉
Γ(H) = •b ##
c
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c
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JU_it~
•b
pp
•
b
@@
c
99tt
tt
tt
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H 6ff H ∗ 〈 〉
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 46 / 69
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Example
H = 〈b, cbc, c2bc−2〉
Γ(H) = •b ## a
**
c
��00
00
00
0 •
c
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JU_it~
•b
pp
•
b
@@
c
99tt
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H 6ff H ∗ 〈ac〉
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 47 / 69
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Example
H = 〈b, cbc, c2bc−2〉
Γ(H) = •b ## a
**
c
��00
00
00
0 •a
jj
c
��@
JU_it~
•b
pp
•
b
@@
c
99tt
tt
tt
tt
tt
t
H 6ff H ∗ 〈ac, c−1a〉
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 48 / 69
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Example
H = 〈b, cbc, c2bc−2〉
Γ(H) = •b ## a
**
c
��00
00
00
0 •a
jj
b
��
c
��@
JU_it~
•b
pp
•
b
@@
c
99tt
tt
tt
tt
tt
t
H 6ff H ∗ 〈ac, c−1a, c−1bc−1〉
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 49 / 69
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Example
H = 〈b, cbc, c2bc−2〉
Γ(H) = •b ## a
**
c
��00
00
00
0 •a
jj
b
��
c
��@
JU_it~
•b
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•
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@@
c
99tt
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tt
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H 6ff H ∗ 〈ac, c−1a, c−1bc−1, c4〉
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 50 / 69
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Example
H = 〈b, cbc, c2bc−2〉
Γ(H) = •b ## a
**
c
��00
00
00
0 •a
jj
b
��
c
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JU_it~
•
a
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c
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H 6ff H ∗ 〈ac, c−1a, c−1bc−1, c4, c2ac−2〉
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 51 / 69
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Example
H = 〈b, cbc, c2bc−2〉
Γ(H) = •b ## a
**
c
��00
00
00
0 •a
jj
b
��
c
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a
NN
H 6ff H ∗ 〈ac, c−1a, c−1bc−1, c4, c2ac−2, cac−1〉 64 F3.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 52 / 69
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Example
H = 〈b, cbc, c2bc−2〉
Γ(H) = •b ## a
**
c
��00
00
00
0 •a
jj
b
��
c
��@
JU_it~
•
a
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•
b
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c
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tt
tt
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tt
t
a
NN
H 6ff H ∗ 〈ac, c−1a, c−1bc−1, c4, c2ac−2, cac−1〉 64 F3.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 52 / 69
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Pull-back of automata
DefinitionThe pull-back of two Stallings automata, (X , v) and (Y , w), is the cartesianproduct (X × Y , (v , w)) (respecting labels). This is not in general connected,neither without degree 1 vertices, but it is folded.
Theorem (H. Neumann-Stallings)For every f.g. subgroups H, K 6fg FA, Γ(H ∩ K ) coincides with the connectedcomponent of Γ(H)× Γ(K ) containing the basepoint, after trimming.
This gives a very nice and quick algorithm to compute intersections:
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 53 / 69
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Pull-back of automata
DefinitionThe pull-back of two Stallings automata, (X , v) and (Y , w), is the cartesianproduct (X × Y , (v , w)) (respecting labels). This is not in general connected,neither without degree 1 vertices, but it is folded.
Theorem (H. Neumann-Stallings)For every f.g. subgroups H, K 6fg FA, Γ(H ∩ K ) coincides with the connectedcomponent of Γ(H)× Γ(K ) containing the basepoint, after trimming.
This gives a very nice and quick algorithm to compute intersections:
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 53 / 69
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Pull-back of automata
DefinitionThe pull-back of two Stallings automata, (X , v) and (Y , w), is the cartesianproduct (X × Y , (v , w)) (respecting labels). This is not in general connected,neither without degree 1 vertices, but it is folded.
Theorem (H. Neumann-Stallings)For every f.g. subgroups H, K 6fg FA, Γ(H ∩ K ) coincides with the connectedcomponent of Γ(H)× Γ(K ) containing the basepoint, after trimming.
This gives a very nice and quick algorithm to compute intersections:
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 53 / 69
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Computing intersections: an example
Let H = 〈a, b2, bab〉 and K = 〈b2, ba2〉 be subgroups of F2.To compute a basis for H ∩ K :
• a // •b
(( •b
hh
a
��
•a$$
b��•a
$$b
GG
H ∩ K =? Clear that b2 ∈ H, but.... something else?
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 54 / 69
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Computing intersections: an example
Let H = 〈a, b2, bab〉 and K = 〈b2, ba2〉 be subgroups of F2.To compute a basis for H ∩ K :
• a // •b
(( •b
hh
a
��
•a$$
b��•a
$$b
GG
H ∩ K =? Clear that b2 ∈ H, but.... something else?
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 54 / 69
![Page 136: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/136.jpg)
Computing intersections: an example
Let H = 〈a, b2, bab〉 and K = 〈b2, ba2〉 be subgroups of F2.To compute a basis for H ∩ K :
• a // •b
(( •b
hh
a
��
•a$$
b��•a
$$b
GG
H ∩ K =? Clear that b2 ∈ H, but.... something else?
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 54 / 69
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Computing intersections: an example
Let H = 〈a, b2, bab〉 and K = 〈b2, ba2〉 be subgroups of F2.To compute a basis for H ∩ K :
• a // •b
(( •b
hh
a
��
•a$$
b��
• • •
•a$$
b
GG
• • •
H ∩ K = 〈b2, . . . (?) . . .〉
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 55 / 69
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Computing intersections: an example
Let H = 〈a, b2, bab〉 and K = 〈b2, ba2〉 be subgroups of F2.To compute a basis for H ∩ K :
• a // •b
(( •b
hh
a
��
•a$$
b��
• •b
��
•
bvv•a$$
b
GG
• •
b66
•b
VV
H ∩ K = 〈b2, 〉
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 56 / 69
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Computing intersections: an example
Let H = 〈a, b2, bab〉 and K = 〈b2, ba2〉 be subgroups of F2.To compute a basis for H ∩ K :
• a // •b
(( •b
hh
a
��
•a$$
b��
• a // •b
��
•
bvv
a��
•a$$
b
GG
• •
b66
•b
VV
H ∩ K = 〈b2, a−2b2a2, 〉
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 57 / 69
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Computing intersections: an example
Let H = 〈a, b2, bab〉 and K = 〈b2, ba2〉 be subgroups of F2.To compute a basis for H ∩ K :
• a // •b
(( •b
hh
a
��
•a$$
b��
• a // •b
��
•
bvv
a��
•a$$
b
GG
• a // •
b66
•b
VV
H ∩ K = 〈b2, a−2b2a2, 〉
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 58 / 69
![Page 141: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/141.jpg)
Computing intersections: an example
Let H = 〈a, b2, bab〉 and K = 〈b2, ba2〉 be subgroups of F2.To compute a basis for H ∩ K :
• a // •b
(( •b
hh
a
��
•a$$
b��
• a // •b
��
•
bvv
a��
•a$$
b
GG
• a // •
b66
•b
VV
a
__
H ∩ K = 〈b2, a−2b2a2, ba2ba2〉 ... and nothing else.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 59 / 69
![Page 142: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/142.jpg)
Computing intersections: an example
Let H = 〈a, b2, bab〉 and K = 〈b2, ba2〉 be subgroups of F2.To compute a basis for H ∩ K :
• a // •b
(( •b
hh
a
��
•a$$
b��
• a // •b
��
•
bvv
a��
•a$$
b
GG
• a // •
b66
•b
VV
a
__
H ∩ K = 〈b2, a−2b2a2, ba2ba2〉 ... and nothing else.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 59 / 69
![Page 143: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/143.jpg)
Rank of the intersection
Theorem (Howson)The intersection of finitely generated subgroups of FA is again finitelygenerated.
But the intersection can have bigger rank: “3 = 3 ∩ 2 6 2”
Theorem (H. Neumann)r̃(H ∩ K ) 6 2r̃(H)r̃(K ), where r̃(H) = max{0, r(H)− 1}.
Conjecture (H. Neumann)r̃(H ∩ K ) 6 r̃(H)r̃(K ).
In the example, 3− 1 6 (3− 1)(2− 1).Enric Ventura (UPC) Automata and Group Theory November 25, 2008 60 / 69
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Rank of the intersection
Theorem (Howson)The intersection of finitely generated subgroups of FA is again finitelygenerated.
But the intersection can have bigger rank: “3 = 3 ∩ 2 6 2”
Theorem (H. Neumann)r̃(H ∩ K ) 6 2r̃(H)r̃(K ), where r̃(H) = max{0, r(H)− 1}.
Conjecture (H. Neumann)r̃(H ∩ K ) 6 r̃(H)r̃(K ).
In the example, 3− 1 6 (3− 1)(2− 1).Enric Ventura (UPC) Automata and Group Theory November 25, 2008 60 / 69
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Rank of the intersection
Theorem (Howson)The intersection of finitely generated subgroups of FA is again finitelygenerated.
But the intersection can have bigger rank: “3 = 3 ∩ 2 6 2”
Theorem (H. Neumann)r̃(H ∩ K ) 6 2r̃(H)r̃(K ), where r̃(H) = max{0, r(H)− 1}.
Conjecture (H. Neumann)r̃(H ∩ K ) 6 r̃(H)r̃(K ).
In the example, 3− 1 6 (3− 1)(2− 1).Enric Ventura (UPC) Automata and Group Theory November 25, 2008 60 / 69
![Page 146: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/146.jpg)
Rank of the intersection
Theorem (Howson)The intersection of finitely generated subgroups of FA is again finitelygenerated.
But the intersection can have bigger rank: “3 = 3 ∩ 2 6 2”
Theorem (H. Neumann)r̃(H ∩ K ) 6 2r̃(H)r̃(K ), where r̃(H) = max{0, r(H)− 1}.
Conjecture (H. Neumann)r̃(H ∩ K ) 6 r̃(H)r̃(K ).
In the example, 3− 1 6 (3− 1)(2− 1).Enric Ventura (UPC) Automata and Group Theory November 25, 2008 60 / 69
![Page 147: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/147.jpg)
Rank of the intersection
Theorem (Howson)The intersection of finitely generated subgroups of FA is again finitelygenerated.
But the intersection can have bigger rank: “3 = 3 ∩ 2 6 2”
Theorem (H. Neumann)r̃(H ∩ K ) 6 2r̃(H)r̃(K ), where r̃(H) = max{0, r(H)− 1}.
Conjecture (H. Neumann)r̃(H ∩ K ) 6 r̃(H)r̃(K ).
In the example, 3− 1 6 (3− 1)(2− 1).Enric Ventura (UPC) Automata and Group Theory November 25, 2008 60 / 69
![Page 148: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/148.jpg)
Status of Hanna Neumann Conjecture
HNC holds if H (or K ) has rank 1 (immediate),
HNC holds for finite index subgroups (elementary),
HNC holds if H has rank 2 (Tardös, 1992), (not easy),
HNC holds if H has rank 3 (Dicks-Formanek, 2001), (quite difficult),
HNC also holds if H is positively generated (⇔ Γ(H) is stronglyconnected), (Meakin-Weil, and Khan, 2002),
HNC in general is an open problem (...and considered very hard).
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 61 / 69
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Status of Hanna Neumann Conjecture
HNC holds if H (or K ) has rank 1 (immediate),
HNC holds for finite index subgroups (elementary),
HNC holds if H has rank 2 (Tardös, 1992), (not easy),
HNC holds if H has rank 3 (Dicks-Formanek, 2001), (quite difficult),
HNC also holds if H is positively generated (⇔ Γ(H) is stronglyconnected), (Meakin-Weil, and Khan, 2002),
HNC in general is an open problem (...and considered very hard).
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 61 / 69
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Status of Hanna Neumann Conjecture
HNC holds if H (or K ) has rank 1 (immediate),
HNC holds for finite index subgroups (elementary),
HNC holds if H has rank 2 (Tardös, 1992), (not easy),
HNC holds if H has rank 3 (Dicks-Formanek, 2001), (quite difficult),
HNC also holds if H is positively generated (⇔ Γ(H) is stronglyconnected), (Meakin-Weil, and Khan, 2002),
HNC in general is an open problem (...and considered very hard).
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 61 / 69
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Status of Hanna Neumann Conjecture
HNC holds if H (or K ) has rank 1 (immediate),
HNC holds for finite index subgroups (elementary),
HNC holds if H has rank 2 (Tardös, 1992), (not easy),
HNC holds if H has rank 3 (Dicks-Formanek, 2001), (quite difficult),
HNC also holds if H is positively generated (⇔ Γ(H) is stronglyconnected), (Meakin-Weil, and Khan, 2002),
HNC in general is an open problem (...and considered very hard).
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 61 / 69
![Page 152: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/152.jpg)
Status of Hanna Neumann Conjecture
HNC holds if H (or K ) has rank 1 (immediate),
HNC holds for finite index subgroups (elementary),
HNC holds if H has rank 2 (Tardös, 1992), (not easy),
HNC holds if H has rank 3 (Dicks-Formanek, 2001), (quite difficult),
HNC also holds if H is positively generated (⇔ Γ(H) is stronglyconnected), (Meakin-Weil, and Khan, 2002),
HNC in general is an open problem (...and considered very hard).
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 61 / 69
![Page 153: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/153.jpg)
Status of Hanna Neumann Conjecture
HNC holds if H (or K ) has rank 1 (immediate),
HNC holds for finite index subgroups (elementary),
HNC holds if H has rank 2 (Tardös, 1992), (not easy),
HNC holds if H has rank 3 (Dicks-Formanek, 2001), (quite difficult),
HNC also holds if H is positively generated (⇔ Γ(H) is stronglyconnected), (Meakin-Weil, and Khan, 2002),
HNC in general is an open problem (...and considered very hard).
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 61 / 69
![Page 154: Automata and Group Theory - UPC Universitat Politècnica de …epsem.upc.edu/~ventura/ventura/files/Bratislava-25-11-08.pdf · 2008. 11. 25. · 1The friendly and unfriendly free](https://reader036.fdocuments.co/reader036/viewer/2022071506/612724460bf857642b6f730c/html5/thumbnails/154.jpg)
Outline
1 The friendly and unfriendly free group
2 The bijection between subgroups and automata
3 Several algorithmic applications
4 Algebraic extensions and Takahasi’s theorem
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 62 / 69
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Free and algebraic extensions
DefinitionAnd extension of subgroups H 6 K , in FA is called
a free extension if H is a free factor of K (i.e. K = H ∗ L for some L 6 FA),denoted H 6ff K ;algebraic if H is not contained in any proper free factor of K (i.e.H 6 K1 6 K1 ∗ K2 = K implies K2 = 1), denoted H 6alg K .
〈a〉 6ff 〈a, b〉 6ff 〈a, b, c〉, and 〈x r 〉 6alg 〈x〉, ∀x ∈ FA ∀r ∈ Z.if r(H) > 2 and r(K ) 6 2 then H 6alg K .H 6alg K 6alg L implies H 6alg L.H 6ff K 6ff L implies H 6ff L.H 6alg L and H 6 K 6 L imply K 6alg L but not necessarily H 6alg K .H 6ff L and H 6 K 6 L imply H 6ff K but not necessarily K 6ff L.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 63 / 69
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Free and algebraic extensions
DefinitionAnd extension of subgroups H 6 K , in FA is called
a free extension if H is a free factor of K (i.e. K = H ∗ L for some L 6 FA),denoted H 6ff K ;algebraic if H is not contained in any proper free factor of K (i.e.H 6 K1 6 K1 ∗ K2 = K implies K2 = 1), denoted H 6alg K .
〈a〉 6ff 〈a, b〉 6ff 〈a, b, c〉, and 〈x r 〉 6alg 〈x〉, ∀x ∈ FA ∀r ∈ Z.if r(H) > 2 and r(K ) 6 2 then H 6alg K .H 6alg K 6alg L implies H 6alg L.H 6ff K 6ff L implies H 6ff L.H 6alg L and H 6 K 6 L imply K 6alg L but not necessarily H 6alg K .H 6ff L and H 6 K 6 L imply H 6ff K but not necessarily K 6ff L.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 63 / 69
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Free and algebraic extensions
DefinitionAnd extension of subgroups H 6 K , in FA is called
a free extension if H is a free factor of K (i.e. K = H ∗ L for some L 6 FA),denoted H 6ff K ;algebraic if H is not contained in any proper free factor of K (i.e.H 6 K1 6 K1 ∗ K2 = K implies K2 = 1), denoted H 6alg K .
〈a〉 6ff 〈a, b〉 6ff 〈a, b, c〉, and 〈x r 〉 6alg 〈x〉, ∀x ∈ FA ∀r ∈ Z.if r(H) > 2 and r(K ) 6 2 then H 6alg K .H 6alg K 6alg L implies H 6alg L.H 6ff K 6ff L implies H 6ff L.H 6alg L and H 6 K 6 L imply K 6alg L but not necessarily H 6alg K .H 6ff L and H 6 K 6 L imply H 6ff K but not necessarily K 6ff L.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 63 / 69
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Free and algebraic extensions
DefinitionAnd extension of subgroups H 6 K , in FA is called
a free extension if H is a free factor of K (i.e. K = H ∗ L for some L 6 FA),denoted H 6ff K ;algebraic if H is not contained in any proper free factor of K (i.e.H 6 K1 6 K1 ∗ K2 = K implies K2 = 1), denoted H 6alg K .
〈a〉 6ff 〈a, b〉 6ff 〈a, b, c〉, and 〈x r 〉 6alg 〈x〉, ∀x ∈ FA ∀r ∈ Z.if r(H) > 2 and r(K ) 6 2 then H 6alg K .H 6alg K 6alg L implies H 6alg L.H 6ff K 6ff L implies H 6ff L.H 6alg L and H 6 K 6 L imply K 6alg L but not necessarily H 6alg K .H 6ff L and H 6 K 6 L imply H 6ff K but not necessarily K 6ff L.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 63 / 69
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Free and algebraic extensions
DefinitionAnd extension of subgroups H 6 K , in FA is called
a free extension if H is a free factor of K (i.e. K = H ∗ L for some L 6 FA),denoted H 6ff K ;algebraic if H is not contained in any proper free factor of K (i.e.H 6 K1 6 K1 ∗ K2 = K implies K2 = 1), denoted H 6alg K .
〈a〉 6ff 〈a, b〉 6ff 〈a, b, c〉, and 〈x r 〉 6alg 〈x〉, ∀x ∈ FA ∀r ∈ Z.if r(H) > 2 and r(K ) 6 2 then H 6alg K .H 6alg K 6alg L implies H 6alg L.H 6ff K 6ff L implies H 6ff L.H 6alg L and H 6 K 6 L imply K 6alg L but not necessarily H 6alg K .H 6ff L and H 6 K 6 L imply H 6ff K but not necessarily K 6ff L.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 63 / 69
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Free and algebraic extensions
DefinitionAnd extension of subgroups H 6 K , in FA is called
a free extension if H is a free factor of K (i.e. K = H ∗ L for some L 6 FA),denoted H 6ff K ;algebraic if H is not contained in any proper free factor of K (i.e.H 6 K1 6 K1 ∗ K2 = K implies K2 = 1), denoted H 6alg K .
〈a〉 6ff 〈a, b〉 6ff 〈a, b, c〉, and 〈x r 〉 6alg 〈x〉, ∀x ∈ FA ∀r ∈ Z.if r(H) > 2 and r(K ) 6 2 then H 6alg K .H 6alg K 6alg L implies H 6alg L.H 6ff K 6ff L implies H 6ff L.H 6alg L and H 6 K 6 L imply K 6alg L but not necessarily H 6alg K .H 6ff L and H 6 K 6 L imply H 6ff K but not necessarily K 6ff L.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 63 / 69
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Free and algebraic extensions
DefinitionAnd extension of subgroups H 6 K , in FA is called
a free extension if H is a free factor of K (i.e. K = H ∗ L for some L 6 FA),denoted H 6ff K ;algebraic if H is not contained in any proper free factor of K (i.e.H 6 K1 6 K1 ∗ K2 = K implies K2 = 1), denoted H 6alg K .
〈a〉 6ff 〈a, b〉 6ff 〈a, b, c〉, and 〈x r 〉 6alg 〈x〉, ∀x ∈ FA ∀r ∈ Z.if r(H) > 2 and r(K ) 6 2 then H 6alg K .H 6alg K 6alg L implies H 6alg L.H 6ff K 6ff L implies H 6ff L.H 6alg L and H 6 K 6 L imply K 6alg L but not necessarily H 6alg K .H 6ff L and H 6 K 6 L imply H 6ff K but not necessarily K 6ff L.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 63 / 69
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Free and algebraic extensions
DefinitionAnd extension of subgroups H 6 K , in FA is called
a free extension if H is a free factor of K (i.e. K = H ∗ L for some L 6 FA),denoted H 6ff K ;algebraic if H is not contained in any proper free factor of K (i.e.H 6 K1 6 K1 ∗ K2 = K implies K2 = 1), denoted H 6alg K .
〈a〉 6ff 〈a, b〉 6ff 〈a, b, c〉, and 〈x r 〉 6alg 〈x〉, ∀x ∈ FA ∀r ∈ Z.if r(H) > 2 and r(K ) 6 2 then H 6alg K .H 6alg K 6alg L implies H 6alg L.H 6ff K 6ff L implies H 6ff L.H 6alg L and H 6 K 6 L imply K 6alg L but not necessarily H 6alg K .H 6ff L and H 6 K 6 L imply H 6ff K but not necessarily K 6ff L.
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 63 / 69
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Takahasi’s Theorem
Theorem (Takahasi, 1951)For every H 6fg FA, the set of algebraic extensions, denoted AE(H), is finite.
Original proof by Takahasi was combinatorial and technical,
Modern proof, using Stallings automata, is much simpler, and dueindependently to Ventura (1997), Margolis-Sapir-Weil (2001) andKapovich-Miasnikov (2002).
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 64 / 69
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Takahasi’s Theorem
Theorem (Takahasi, 1951)For every H 6fg FA, the set of algebraic extensions, denoted AE(H), is finite.
Original proof by Takahasi was combinatorial and technical,
Modern proof, using Stallings automata, is much simpler, and dueindependently to Ventura (1997), Margolis-Sapir-Weil (2001) andKapovich-Miasnikov (2002).
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 64 / 69
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Takahasi’s Theorem
Theorem (Takahasi, 1951)For every H 6fg FA, the set of algebraic extensions, denoted AE(H), is finite.
Original proof by Takahasi was combinatorial and technical,
Modern proof, using Stallings automata, is much simpler, and dueindependently to Ventura (1997), Margolis-Sapir-Weil (2001) andKapovich-Miasnikov (2002).
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 64 / 69
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The modern proof
Proof:Let us (temporarily) attach some “hairs" to Γ(H) an denote the resulting(folded) automata by Γ̃(H).Given H 6 K (both f.g.), we can obtain Γ(K ) from Γ(H) by 1) adding theappropriate hairs, 2) identifying several vertices to •, 3) folding; (note thatadding extra hairs, the result will be the same if we 4) trim at the end).Hence, if H 6 K (both f.g.) then Γ(K ) contains as a subgraph either Γ(H)or some quotient of it (i.e. Γ(H) after identifying several sets of vertices(∼) and then folding, Γ(H)/ ∼).The overgroups of H:O(H) = {π(Γ(H)/ ∼, •) | ∼ is a partition of VΓ(H)}.Hence, for every H 6 K , there exists L ∈ O(H) such that H 6 L 6ff K .Thus, AE(H) ⊆ O(H) and so, it is finite. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 65 / 69
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The modern proof
Proof:Let us (temporarily) attach some “hairs" to Γ(H) an denote the resulting(folded) automata by Γ̃(H).Given H 6 K (both f.g.), we can obtain Γ(K ) from Γ(H) by 1) adding theappropriate hairs, 2) identifying several vertices to •, 3) folding; (note thatadding extra hairs, the result will be the same if we 4) trim at the end).Hence, if H 6 K (both f.g.) then Γ(K ) contains as a subgraph either Γ(H)or some quotient of it (i.e. Γ(H) after identifying several sets of vertices(∼) and then folding, Γ(H)/ ∼).The overgroups of H:O(H) = {π(Γ(H)/ ∼, •) | ∼ is a partition of VΓ(H)}.Hence, for every H 6 K , there exists L ∈ O(H) such that H 6 L 6ff K .Thus, AE(H) ⊆ O(H) and so, it is finite. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 65 / 69
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The modern proof
Proof:Let us (temporarily) attach some “hairs" to Γ(H) an denote the resulting(folded) automata by Γ̃(H).Given H 6 K (both f.g.), we can obtain Γ(K ) from Γ(H) by 1) adding theappropriate hairs, 2) identifying several vertices to •, 3) folding; (note thatadding extra hairs, the result will be the same if we 4) trim at the end).Hence, if H 6 K (both f.g.) then Γ(K ) contains as a subgraph either Γ(H)or some quotient of it (i.e. Γ(H) after identifying several sets of vertices(∼) and then folding, Γ(H)/ ∼).The overgroups of H:O(H) = {π(Γ(H)/ ∼, •) | ∼ is a partition of VΓ(H)}.Hence, for every H 6 K , there exists L ∈ O(H) such that H 6 L 6ff K .Thus, AE(H) ⊆ O(H) and so, it is finite. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 65 / 69
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The modern proof
Proof:Let us (temporarily) attach some “hairs" to Γ(H) an denote the resulting(folded) automata by Γ̃(H).Given H 6 K (both f.g.), we can obtain Γ(K ) from Γ(H) by 1) adding theappropriate hairs, 2) identifying several vertices to •, 3) folding; (note thatadding extra hairs, the result will be the same if we 4) trim at the end).Hence, if H 6 K (both f.g.) then Γ(K ) contains as a subgraph either Γ(H)or some quotient of it (i.e. Γ(H) after identifying several sets of vertices(∼) and then folding, Γ(H)/ ∼).The overgroups of H:O(H) = {π(Γ(H)/ ∼, •) | ∼ is a partition of VΓ(H)}.Hence, for every H 6 K , there exists L ∈ O(H) such that H 6 L 6ff K .Thus, AE(H) ⊆ O(H) and so, it is finite. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 65 / 69
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The modern proof
Proof:Let us (temporarily) attach some “hairs" to Γ(H) an denote the resulting(folded) automata by Γ̃(H).Given H 6 K (both f.g.), we can obtain Γ(K ) from Γ(H) by 1) adding theappropriate hairs, 2) identifying several vertices to •, 3) folding; (note thatadding extra hairs, the result will be the same if we 4) trim at the end).Hence, if H 6 K (both f.g.) then Γ(K ) contains as a subgraph either Γ(H)or some quotient of it (i.e. Γ(H) after identifying several sets of vertices(∼) and then folding, Γ(H)/ ∼).The overgroups of H:O(H) = {π(Γ(H)/ ∼, •) | ∼ is a partition of VΓ(H)}.Hence, for every H 6 K , there exists L ∈ O(H) such that H 6 L 6ff K .Thus, AE(H) ⊆ O(H) and so, it is finite. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 65 / 69
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The modern proof
Proof:Let us (temporarily) attach some “hairs" to Γ(H) an denote the resulting(folded) automata by Γ̃(H).Given H 6 K (both f.g.), we can obtain Γ(K ) from Γ(H) by 1) adding theappropriate hairs, 2) identifying several vertices to •, 3) folding; (note thatadding extra hairs, the result will be the same if we 4) trim at the end).Hence, if H 6 K (both f.g.) then Γ(K ) contains as a subgraph either Γ(H)or some quotient of it (i.e. Γ(H) after identifying several sets of vertices(∼) and then folding, Γ(H)/ ∼).The overgroups of H:O(H) = {π(Γ(H)/ ∼, •) | ∼ is a partition of VΓ(H)}.Hence, for every H 6 K , there exists L ∈ O(H) such that H 6 L 6ff K .Thus, AE(H) ⊆ O(H) and so, it is finite. �
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 65 / 69
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Computing AE(H)
CorollaryAE(H) is computable.
Proof:Compute Γ(H),Compute Γ(H)/ ∼ for all partitions ∼ of VΓ(H),Compute O(H),Clean O(H) by detecting all pairs K1, K2 ∈ O(H) such that K1 6ff K2 anddeleting K2.The resulting set is AE(H). �
For the cleaning step we need:
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 66 / 69
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Computing AE(H)
CorollaryAE(H) is computable.
Proof:Compute Γ(H),Compute Γ(H)/ ∼ for all partitions ∼ of VΓ(H),Compute O(H),Clean O(H) by detecting all pairs K1, K2 ∈ O(H) such that K1 6ff K2 anddeleting K2.The resulting set is AE(H). �
For the cleaning step we need:
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 66 / 69
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Computing AE(H)
CorollaryAE(H) is computable.
Proof:Compute Γ(H),Compute Γ(H)/ ∼ for all partitions ∼ of VΓ(H),Compute O(H),Clean O(H) by detecting all pairs K1, K2 ∈ O(H) such that K1 6ff K2 anddeleting K2.The resulting set is AE(H). �
For the cleaning step we need:
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 66 / 69
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Computing AE(H)
CorollaryAE(H) is computable.
Proof:Compute Γ(H),Compute Γ(H)/ ∼ for all partitions ∼ of VΓ(H),Compute O(H),Clean O(H) by detecting all pairs K1, K2 ∈ O(H) such that K1 6ff K2 anddeleting K2.The resulting set is AE(H). �
For the cleaning step we need:
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 66 / 69
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Computing AE(H)
CorollaryAE(H) is computable.
Proof:Compute Γ(H),Compute Γ(H)/ ∼ for all partitions ∼ of VΓ(H),Compute O(H),Clean O(H) by detecting all pairs K1, K2 ∈ O(H) such that K1 6ff K2 anddeleting K2.The resulting set is AE(H). �
For the cleaning step we need:
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 66 / 69
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Computing AE(H)
CorollaryAE(H) is computable.
Proof:Compute Γ(H),Compute Γ(H)/ ∼ for all partitions ∼ of VΓ(H),Compute O(H),Clean O(H) by detecting all pairs K1, K2 ∈ O(H) such that K1 6ff K2 anddeleting K2.The resulting set is AE(H). �
For the cleaning step we need:
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 66 / 69
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Computing AE(H)
CorollaryAE(H) is computable.
Proof:Compute Γ(H),Compute Γ(H)/ ∼ for all partitions ∼ of VΓ(H),Compute O(H),Clean O(H) by detecting all pairs K1, K2 ∈ O(H) such that K1 6ff K2 anddeleting K2.The resulting set is AE(H). �
For the cleaning step we need:
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 66 / 69
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Deciding free-factorness
PropositionGiven H, K 6 FA, it is algorithmically decidable whether H 6ff K or not.
Proved by:Whitehead 1930’s (classical and exponential),Silva-Weil 2006 (graphical algorithm, faster but still exponential),Roig-Ventura-Weil 2007 (variation of Whitehead algorithm in polynomialtime).
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 67 / 69
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Deciding free-factorness
PropositionGiven H, K 6 FA, it is algorithmically decidable whether H 6ff K or not.
Proved by:Whitehead 1930’s (classical and exponential),Silva-Weil 2006 (graphical algorithm, faster but still exponential),Roig-Ventura-Weil 2007 (variation of Whitehead algorithm in polynomialtime).
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 67 / 69
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Deciding free-factorness
PropositionGiven H, K 6 FA, it is algorithmically decidable whether H 6ff K or not.
Proved by:Whitehead 1930’s (classical and exponential),Silva-Weil 2006 (graphical algorithm, faster but still exponential),Roig-Ventura-Weil 2007 (variation of Whitehead algorithm in polynomialtime).
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 67 / 69
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Deciding free-factorness
PropositionGiven H, K 6 FA, it is algorithmically decidable whether H 6ff K or not.
Proved by:Whitehead 1930’s (classical and exponential),Silva-Weil 2006 (graphical algorithm, faster but still exponential),Roig-Ventura-Weil 2007 (variation of Whitehead algorithm in polynomialtime).
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 67 / 69
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The algebraic closure
ObservationIf H 6alg K1 and H 6alg K2 then H 6alg 〈K1 ∪ K2〉.
CorollaryFor every H 6 K 6 FA (all f.g.), AEK (H) has a unique maximal element, calledthe K -algebraic closure of H, and denoted ClK (H).
CorollaryEvery extension H 6 K of f.g. subgroups of FA splits, in a unique way, in analgebraic part and a free factor part, H 6alg Cl(H) 6ff K .
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 68 / 69
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The algebraic closure
ObservationIf H 6alg K1 and H 6alg K2 then H 6alg 〈K1 ∪ K2〉.
CorollaryFor every H 6 K 6 FA (all f.g.), AEK (H) has a unique maximal element, calledthe K -algebraic closure of H, and denoted ClK (H).
CorollaryEvery extension H 6 K of f.g. subgroups of FA splits, in a unique way, in analgebraic part and a free factor part, H 6alg Cl(H) 6ff K .
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 68 / 69
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The algebraic closure
ObservationIf H 6alg K1 and H 6alg K2 then H 6alg 〈K1 ∪ K2〉.
CorollaryFor every H 6 K 6 FA (all f.g.), AEK (H) has a unique maximal element, calledthe K -algebraic closure of H, and denoted ClK (H).
CorollaryEvery extension H 6 K of f.g. subgroups of FA splits, in a unique way, in analgebraic part and a free factor part, H 6alg Cl(H) 6ff K .
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 68 / 69
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THANKS
Enric Ventura (UPC) Automata and Group Theory November 25, 2008 69 / 69