# A finite group such that every element is conjugate to its square is trivial

Suppose $G$ is a finite group such that $g$ is conjugate to $g^2$ for every $g\in G$.

Here's a proof that $G$ is trivial. First, observe that if $\lvert G\rvert$ is even, then $G$ contains an element $h$ of order $2$, in which case, $h$ is conjugate to $h^2=1$. But this implies that $h=1$, so $h$ does not have order $2$. By contradiction, $\lvert G\rvert$ is odd. Then, by the Feit–Thompson theorem, $G$ is solvable. In particular, this means that the derived series of $G$ terminates. However, for any $g$ in $G$, there exists $a\in G$ such that $g^2=aga^{-1}$, i.e., $g=aga^{-1}g^{-1}\in G^{(1)}$. It follows that $G^{(1)}=G$. In fact, this shows that $G^{(n)}=G$ for all $n\geq 1$. Since the derived series of $G$ terminates, this implies that $G$ must be trivial.

While I'm convinced of the result, this proof is not particularly satisfying to me, since it relies on Feit-Thompson. Is there an elementary proof that $G$ is trivial?

• This is a surprisingly hard question (althoughI know very little about finite group theory!) – Andres Mejia Aug 16 '18 at 4:17

As you say, $G$ must have odd order. Let $p$ be the smallest prime factor of the order of $G$, and $a$ an element of order $p$. Let $H$ be the subgroup generated by $a$, $C$ be the centraliser of $H$ and $N$ the normaliser of $H$. Then $r=|N:C|$ is the number of elements of $H$ which are conjugates of $a$. So $r<p$ but $r>1$, as $a^2\ne a$ is the conjugate of $a$. But $r\mid |G|$ and $1<r<p$, contradicting $p$ being the smallest (prime) factor of $|G|$.
• Sorry, why is $r$ the number of elements which are conjugates of $a$? – user614287 Dec 14 '18 at 14:05