Let $G$ be a finite group and let $N\trianglelefteq G$, $2\mid |N|$. If the non-trivial elements of $N$ form a single conjugacy class of $G$, prove that $N$ is abelian.

My Attempt

I tried to approach by using the Orbit-Stabilizer theorem as follows: let $a\in N$ be given, by the Orbit-Stabilizer theorem (the action being $G$ acting on itself by conjugation), we have $$|G| = |O_a||G_a|,$$ where $O_a, G_a$ are the orbit and stabilizer of $a$, respectively.

By assumption, we then have $|G| = (|N|-1)|G_a|$ and we know that $|N|-1$ is odd. Then I am trying to argue that $N\subseteq G_a$, which would prove that $N$ is abelian since the choice of $a$ is arbitrary. But I couldn't make the connection there.

Also, this approach may be totally wrong. But at the moment I couldn't see any other possible way of proving this.

Any help would be greatly appreciated. Thanks.


Well actually, by Cauchy's theorem, $N$ has an element of order $2$. By conjugacy it follows that every nonidentity element is of order $2$. It's a very common exercise that I'm sure you've done to show that if every nonidentity element of a group is of order $2$, then the group is abelian. This actually doesn't require the hypothesis that the subgroup is normal.

  • $\begingroup$ Really nice argument! Thanks a lot, Matt! $\endgroup$ – mkmlp Apr 28 at 16:19
  • 1
    $\begingroup$ @mkmlp No problem. $\endgroup$ – Matt Samuel Apr 28 at 16:20

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