Order of an element of group

[Grigorios Kostakos]

Let \(G\) 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 $b$.

[Tolaso J Kos]

Let \(G\) 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 $b$.

Alright, here is an approach. Hope I did not make any mistakes here.

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 have that $a^nba^{-n}=b^{2^n}$ and thus $b^{31}=e$. Noting that $31$ is prime we deduce that the order of $b$ is $31$ or $1$.

[Grigorios Kostakos]

Another solution:

For the elements \(a,\,b\) of \(G\) we have that:
\begin{align*}
a^5&=e&(1)\\
a\,b&=b^2a&(2)
\end{align*} So \begin{align*}
b\stackrel{(1)\,(2)}{=\!=\!=}a^4b^2a\quad&\Longrightarrow\quad b^{\rm{n}}=\mathop{\underbrace{a^4b^2a\,a^4b^2a\cdots a^4b^2a}}\limits_{{\rm{n}}-\text{times}}\stackrel{(1)}{=}a^4b^{2{\rm{n}}}a\\
&\stackrel{(1)}{\Longrightarrow}\quad b^{2{\rm{n}}}=a\,b^{{\rm{n}}}a^4 \quad(3)\,, \quad{\rm{n}}\in\mathbb{N}\,.
\end{align*} But then
\begin{align*}
b^{32}&\stackrel{(3)\,{\rm{n}}=16}{=\!=\!=\!=}a\,b^{16}a^4\\
&\stackrel{(3)\,{\rm{n}}=8}{=\!=\!=\!=}a\,a\,b^{8}a^4a^4\\
&\stackrel{(3)\,{\rm{n}}=4}{=\!=\!=\!=}a\,a\,a\,b^{4}a^4a^4a^4\\
&\stackrel{(3)\,{\rm{n}}=2}{=\!=\!=\!=}a\,a\,a\,a\,b^{2}a^4a^4a^4a^4\\
&\stackrel{(3)\,{\rm{n}}=1}{=\!=\!=\!=}a\,a\,a\,a\,a\,b^{2}a^4a^4a^4a^4a^4\\
&=a^5b\,(a^5)^4\\
&\stackrel{(3)}{=}e\,b\,e^4\\
&=b\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\Longrightarrow\\
b^{31}&=e\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\Longrightarrow\\
\circ(b)&\,\big|\,31\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\Longrightarrow\\
\circ(b)=1\; &\veebar \; \circ(b)=31\,.
\end{align*}