Pretty easy equations of elements in a group Problem
$G$ is a group generated by $a,b\in G$ such that $a^5=e$, $aba^{-1}=b^2$ and $b\ne e$. I want to find the order of $b$.
Attempt
I tried to multiply the second equation from right by $a^{4}$: $$ba^{-1}=a^{4}b^{2}$$ Then $$ba^{-1}=(ba^{2})^{2}=ba^{2}ba^{2}$$
Then by multiplying by $a$ from right and $b^{-1}$ from left we get
$$a^{2}ba^{3}=e$$ Then by multiplying by $a^{-2}$ from the left and $a^{-3}$ from right we get $$b=a^{-5}=(a^{5})^{-1}=e$$ but $b\ne e$, so is it right to say that the order is $\infty$? Or did I do something wrong?

Note that for every $m$ we have
$$ab^ma^{-1} = (aba^{-1})^m = b^{2m}.$$
Thus by induction we have
$$a^kba^{-k} = b^{2^k}\tag{$\ast$}$$
for all $k\in\mathbb{N}$. On the other hand, we have $a^5ba^{-5} = b$ since $a^5 = e$. Putting that and $(\ast)$ together reveals the order of $b$.
