Does $A$ have infinite order in $G= \langle A,B \ |\ B A B^{-1} = A^2 \rangle $?

I have a group (arising from the fundamental group of a manifold) $$G= \langle A,B \ |\ B A B^{-1} = A^2\rangle $$ and

I would like to show that $A$ is an element of infinite order inside $G$.

Notice that in the abelianization ${\rm Ab}(G)\simeq \mathbb{Z}$, the image of $A$ is the identity. Nevertheless, $A$ is a non-trivial element in $G$, indeed we can find a representation $G\to{\rm SL}(2;\mathbb C)$ that sends $A$ to the diagonal matrix ${\rm diag}(e^{i2\pi/3},e^{-i 2 \pi/3})$.

Solution 1:

The comments have already given two solutions; here's a third. (It's not really new, just a disguised version of those arguments, but I like the elementary description it gives.)

Choose $n$ odd and consider the cyclic group $C_n=\langle\rho\rangle$. The map $\phi:C_n\to C_n$; $$\phi(g)=g^2$$ generates a subgroup of $\mathrm{Aut}{(C_n)}$ that is cyclic of order $m$. Now form a semidirect product $$C_n\rtimes C_m$$ where conjugation by the generator of $C_m$ acts as $\phi$. Then $$\tau\rho\tau^{-1}=\phi(\rho)=\rho^2$$ and so the map $\psi:G\to C_n\rtimes C_m$; \begin{gather*} \psi(A)=\rho \\ \psi(B)=\tau \end{gather*} is a homomorphism.

Moreover, $\psi(A)$ has order $n$. Varying $n$, we see that $G$ has quotients in which the image of $A$ has arbitrarily-large order. But quotients can only decrease the order of an element, so $A$ must have infinite order to begin with.