In this article of Bartholdi and Sunik http://arxiv.org/abs/math/0603032, they say that BS(n,n) and BS(n,-n) are automata groups because they are virtually $F_{|n|}\rtimes\mathbb{Z}$ (where $F_{|n|}$ is the free group of rank $|n|$) following an article of Brunner & Sidki. I would like to know how to prove this statement. By advance thank you.
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$\begingroup$ It is not at all clear what you mean by "this statement". Which statement? $\endgroup$– Derek HoltCommented Aug 19, 2015 at 15:44
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$\begingroup$ In fact I meant both statements. sorry for the confusion $\endgroup$– Edgar NdieCommented Aug 22, 2015 at 6:01
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$\begingroup$ Presumably you have looked at the article by Brunner & Sidki? $\endgroup$– Derek HoltCommented Aug 22, 2015 at 8:15
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$\begingroup$ yes I did. In that article, Brunner & Sidki want to prove that $GL_n(\mathbb{Z})$ is an automaton group. But I think it's not the case. Anyway I don't see any link with the fact that $BS(n,n)$ is an automaton group, $F_{|n|}\rtimes\mathbb{Z},$ and the fact $GL_n(\mathbb{Z})$ is an automaton group. The only link (which is false) I see is even if $BS(n,n)$ is a subgroup of $GL_n(\mathbb{Z})$ and $GL_n(\mathbb{Z})$ is an automaton group, it does not make $BS(n,n)$ an automaton group. $\endgroup$– Edgar NdieCommented Aug 22, 2015 at 11:08
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2 Answers
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Begin with $BS(n,n)=\langle t,x\mid tx^nt^{-1}=x^n\rangle$. Let $L$ be the kernel of the homomorphism $BS(n,n)\to\mathbf{Z}$, $t\mapsto 0$, $x\mapsto 1$. Using the action of $BS(n,n)$ on its standard Bass-Serre tree shows that $L$ is free. It is generated by the $x^mtx^{-m}$ when $m$ ranges over $\mathbf{Z}$, but since this sequence is $n$-periodic, $m$ ranging over $\{0,\dots,n-1\}$ is enough. It can be shown that this is a free family but I won't check it since this is enough to show that $L$ is free of rank $\le n$, which is enough for your purposes. Now $x^n$ is central and $L\times\langle x^n\rangle$ has finite index.
For $BS(n,-n)=\langle t,x:tx^nt^{-1}=x^{-n}\rangle$, we need to consider the subgroup $H$ of index 2 kernel of the homomorphism mapping $x$ to $0$ and $t$ to 1 mod 2. It has the presentation, denoting $u=t^2$ and $y=tx^{-1}t^{-1}$: $\langle u,x,y\mid x^n=y^n, tx^nt^{-1}=x^n,ty^nt^{-1}=y^n\rangle$. But the latter is also isomorphic to a subgroup of index 2 in $BS(n,n)$ (kernel of the same homomorphism, but with $y$ defined as $txt^{—1}\in BS(n,n)$ instead). This yield a (free $\times$ infinite cyclic) subgroup of finite index, but rather $F_{2n-1}\times\mathbf{Z}$ than $F_n\times\mathbf{Z}$ (although I don't claim there's no $F_n\times\mathbf{Z}$ of finite index).
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$\begingroup$ Thank you for your answer. Why does this property make $BS(n,n)$ an automaton group? $\endgroup$ Commented Aug 19, 2015 at 15:16
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$\begingroup$ As pointed out by Derek, your question is ambiguous. My guess was that you understand that (1) being virtually $\mathbf{Z}\times F_k$ implies being automata group and that you were asking (2) why $BS(n,\pm n)$ has this virtual property. If you're asking why (1) holds, somebody else could answer better than me. $\endgroup$– YCorCommented Aug 19, 2015 at 21:38
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$\begingroup$ yes I'm also asking why (1) holds $\endgroup$ Commented Aug 22, 2015 at 5:58
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$\begingroup$ I never heard of "standard Bass-Serre tree" of a group? What is it? $\endgroup$ Commented Aug 22, 2015 at 6:02
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$\begingroup$ For a group there's no "standard Bass-Serre tree". But for Baumslag-Solitar groups there's one: indeed they are defined as HNN extension of an infinite cyclic group, and the corresponding Bass-Serre tree is the one I call "standard", the action is with cyclic edge and vertex stabilizers, and in restriction to $L$ the vertex stabilizers are trivial. $\endgroup$– YCorCommented Aug 22, 2015 at 7:54
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Yves's answer explains why $BS(n,\pm n)$ is virtually (a free group) x $\mathbb Z$. In particular, it is linear over $\mathbb Z$, and so may be generated by automata. This is the statement in the article [BS].
