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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?

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    $\begingroup$ This is a surprisingly hard question (althoughI know very little about finite group theory!) $\endgroup$ – Andres Mejia Aug 16 '18 at 4:17
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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 $|G|$.

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  • $\begingroup$ Sorry, why is $r$ the number of elements which are conjugates of $a$? $\endgroup$ – user614287 Dec 14 '18 at 14:05
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Just to add to Lord Shark's answer, if you are curious, see Lemma 5.1 of the paper Gabriel Navarro, The McKay conjecture and Galois automorphisms, Annals of Mathematics, 160 (2004), 1129–1140.

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    $\begingroup$ Very nice little gem. $\endgroup$ – Nicky Hekster Aug 16 '18 at 11:13

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