Order of an element

Let  $G$ be a group and $a\,,b\in G$, such that  $a^5=e$ and $a\,b\,a^{-1}=b^2$, where $e$ is the identity element of $G$. Find the order of $b$.

Solution

We begin by stating a lemma:

Lemma: If $aba^{−1}=b^r$ then $a^nba^{−n}=b^{r^n}$.

Proof:

First, we multiply $a$ and $a^{-1}$ from right and left respectively. Thus, one can see that $a^2ba^{−2}=ab^ra^{−1}$. Since $ab^ra^{−1}=(aba^{−1})^r$, $a^2ba^{−2}=(aba^{−1})^r$. Now, we use the main relation and so $a^2ba^{−2}=b^{r^2}$. By repeating the previous procedure, one can show that $a^nba^{−n}=b^{r^n}$, as desired.

Now using the lemma we see that $ a^nba^{-n}=b^{2^n}$ and thus $b^{31}=e$. Since $31$ is prime we get that the order of $b$ is $31$ or $1$.