# Normal subgroup of even order whose nontrivial elements form a single conjugacy class is abelian

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.

## 1 Answer

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.

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