Finding order of $gag^{-1}$ in $G$ if $a^2=e\in G$ Let $G$ be a group, the order of $G$ is even, let $a \in G$, $a^2=e$ > I need to find the order of $gag^{-1}$ in $G$ **My attempt:** $(gag^{-1})^2=(gag^{-1})(gag^{-1})=ga(g^{-1}g)ag^{-1}=ga(e)ag^{-1}=ga^2g^{-1}=g(e)g^{-1}=e$ $\Rightarrow$ the order of $gag^{-1}=\boxed{\color{blue}2}$ Is it correct? is there other why to solve this? any **hints** please?

I am just restating what others have already said in the comments..

You have two cases. Case 1 is where $a=e$ is the identity in your group. Here the order of $gag^{-1} = gg^{-1} = e$ is $1$.

Case 2 is where $a\
eq e$. Then your calculation above indeed shows that the order is $2$ (because only the identity has order $1$). And, as mentioned below, $gag^{-1}\
eq e$ because if it was: $gag^{-1} = e$, then $a = g^{-1}eg = g^{-1}g = e$ (which is case 1).