Prove that $o(a)=o(gag^{-1})$ Let $G$ be a group and $a\in G$. Prove that $o(a)=o(gag^{-1})$ for every element of order $2$ in $G$. If a be the only element of order $2$ in $G$ deduce that a commutes with every element of $G$ Approach: Let $o(a)=n$, then $a^n=e$. Now \begin{eqnarray} (gag^{-1})^n &=& gag^{-1}gag^{-1}gag^{-1}...gag^{-1} \\\ &=& ga^ng^{-1} \\\ &=& gg^{-1} \hbox{ since $a^n=e$}\\\ &=& e \end{eqnarray} which shows that $o(gag^{-1})\leq n$ i.e $o(gag^{-1})\leq o(a)$. How to show the $o(gag^{-1})= o(a)$ 2nd part Let $a$ be the only element of order 2 then $a^2=e$. Now $(gag^{-1})^2=\cdots=e$. So if $a$ is the only element of order 2 then $a=gag^{-1}$, or $ag=ga$ for all $g\in G$. Hence the 2nd part. Please correct my 1st part. Check the 2nd part also.

**HINT for the 1st part:**

You have already proven that $o(gag^{-1})\leq o(a)$, this **for all** $a,g\in G$. In order to show the other inequality, notice that

$$a=g^{-1}(gag^{-1})g$$

so you can apply the previous case with $g$ 'equal' to $g^{-1}$ and $a$ 'equal' to $gag^{-1}$.

Here you have more details about it:

> We have proven that for all $b,h\in G: o(hbh^{-1})\leq o(b)$. We want to prove that $o(gag^{-1})\geq o(a)$.
>
> Take $b=gag^{-1}$ and $h=g^{-1}$, then, applying what we have just proven we have $$o(hbh^{-1})=o(g^{-1}gag^{-1}(g^{-1})^{-1})=o(a)\leq o(b)=o(gag^{-1})$$

Your proof of the second part works perfectly, moreover, you can simply omit the reasoning $(gag^{-1})^2=\cdots=e$ since this is exactly what you've done in part 1.