$G=\langle a,b \mid baba^{-1}=1\rangle$ Show that $\langle a \rangle$ is infinite 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.
 A: 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 $z^n = f(x^n) = h(g(x^n)) = h(g(x^m)) = f(x^m) = z^m$).
A: 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. 
A: 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!
