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Let $G=\langle a,b \mid baba^{-1}=1\rangle$. Show that the subgroup generated by $a$ is infinite.

My attempt

Suppose $\langle a\rangle$ is finite so $a^k = 1$ for some $k \in \mathbb{Z}$. So I would like to prove that the relation $a^k=1$ is not a consequence of the original relation $baba^{-1}=1$. I also have been able to prove that every element $g$ in $G$ can be written as $g= a^n b^m$ for $n,m \in \mathbb{Z} $

Any hint will be appreciated.

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Hint: consider the homomorphism $f$ from the free group $F = \langle x, y \rangle$ to the free group $H = \langle z \rangle$ that maps $x$ to $z$ and $y$ to $1$. The element $yxyx^{-1}$ lies in the kernel of $f$, so $f$ factors as $h \circ g$, where $g : F \to G$ and $h : G \to H$ are homomorphisms with $g(x) = a$, $g(y) = b$, $h(a) = z$ and $h(b) = 1$. The elements $f(x^n) = z^n$ for $n \in \Bbb{Z}$ are all distinct, hence the elements $g(x^n) = g(x)^n = a^n$ are also all distinct (if $g(x^n) = g(x^m)$, we have have $z^n = f(x^n) = h(g(x^n)) = h(g(x^m)) = f(x^m) = z^m$).

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Hint. Kill $b$.

(That is, consider what happens when you add in the relator $b=1$ to the presentation. In fact, it relatively is easy to see that the subgroup generated by $b$ is normal and that this group is simply the semidirect product $\mathbb{Z}\rtimes_{\phi}\mathbb{Z}$, where the action $\phi$ corresponds to the non-trivial automorphism of $\mathbb{Z}$.)

In fact, suppose $G$ is given by a presentation of the form $\langle \mathbf{x}\mid R^n\rangle$ where $n\in\mathbb{Z}$ is maximal (so $R$ is not a proper power of any element of $F(\mathbf{x})$). Then $G$ is torsion-free if and only if $n=1$ [see Magnus, Karrass and Solitar, Combinatorial group theory; the section on one-relator groups]. This is the case here; so every non-trivial element of your group (and not just $a$) generates an infinite group!

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    $\begingroup$ I am fond of the answers who are older in this site. Concise hint and effective! $\endgroup$ – mrs Sep 11 '18 at 7:23
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I think you are shooting in the wrong direction. Try to find arbitrarily large finite groups generated by two elements that satisfy this defining relation. Hint: look for commutative groups! Then use von Dyck's theorem (see http://mathworld.wolfram.com/vonDycksTheorem.html ) to conclude that in the group you presented the order of $a$ must be infinite.

In fact, it is also not so hard to find an infinite (commutative) group generated by at most two elements that satisfy this defining relation, such that the generator corresponding to $a$ has infinite order.

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  • $\begingroup$ Why the negative vote? $\endgroup$ – A. Pongrácz Sep 9 '18 at 20:19
  • $\begingroup$ What you write is not incorrect, but you seem to be proving that the group is infinite, whereas the exercise was to show that the element $a$ has infinite order. $\endgroup$ – Derek Holt Sep 9 '18 at 20:41
  • $\begingroup$ You are right. But in fact, this strategy also works for that. Thank you for pointing out this oversight. $\endgroup$ – A. Pongrácz Sep 9 '18 at 20:42
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    $\begingroup$ I didn't downvote your answer (and I know how annoying it is to be downvoted with no explanation). The bit in your answer about arbitrarily large finite groups seems to me to be pointing in an irrelevant direction, so maybe that was the reason for the downvote. $\endgroup$ – Rob Arthan Sep 9 '18 at 21:41
  • $\begingroup$ @Rob Arthan Of course it is not necessary. But I prefer adding some extra perspective to my answers, rather than providing a direct proof. There are some problems where it is easier to find arbitrarily large finite images than an infinite one. Sometimes you find the infinite image after you understand the finite ones. In any case, the answer is perfectly correct, and all the rest is a matter of taste. Not upvoting: sure, if you do not like it. But downvoting a correct answer? That is way too harsh. Anyway, thank you for your suggestion. $\endgroup$ – A. Pongrácz Sep 10 '18 at 12:53

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