TITLE: Is this reasoning correct?
QUESTION [5 upvotes]: Let G be a finite group of uneven order. Let $x \in G$ be an element with the property that there exist an element $g \in  G$ such that $gxg^{-1} = x^{-1}$. Prove $x$ is the idendity of $G$.
Here is my reasoning, but I think it is incorrect. Why ?
\begin{align*}
x^{-1}=gxg^{-1} &\Rightarrow gx = x^{-1}g \\
&\Rightarrow (gx)^2 = (gx)(x^{-1}g)=g^2 \\ 
&\Rightarrow gx =g \\
&\Rightarrow x =e
\end{align*}

REPLY [2 votes]: Almost complete! The gap in the proof  is when you go from  from $(gx)^2=g^2$ to $gx=g$. Here you need to use the fact that the group is of odd order. 
Note that the order of an element must divide the order of the group. So a group of odd order has no elements of order $2$. Use that tool to finish. 
Added: We need to show that any square has a unique "square root." Consider the object $a^2$. It Since the order of $a^2$ divides the order of the group, $a^2$ has odd order, say $2k+1$. So $a^{2k+1}=e$, and therefore $a^{2k+2}=a$.  So $a=(a^2)^{k+1}$. In particular, $a$ is completely determined by $a^2$.