If a group has an automorphism which is fixed point free of order 2, then it is abelian

Let $G$ be a finite group which possesses an automorphism $\sigma$ such that $\sigma(g) = g$ if and only if $g = 1$. If $\sigma^2 = \sigma \circ \sigma$ is the identity map on $G$, prove that $G$ is abelian. Such an automorphism $\sigma$ is called fixed point free of order 2. [Hint: Show that every element of $G$ can be written in the form $x^{-1} \sigma(x)$ and apply $\sigma$ to such an expression.]

We define a mapping $f : G \rightarrow G$ by $f(x) = x^{-1} \sigma(x)$.

Claim: $f$ is injective.
Proof of claim: Suppose $f(x) = f(y)$. Then $y^{-1} \sigma(y) = x^{-1} \sigma(x)$, so that $xy^{-1} = \sigma(x)\sigma(y^{-1})$, and $xy^{-1} = \sigma(xy^{-1})$. Then we have $xy^{-1} = 1$, hence $x = y$. So $f$ is injective.

Since $G$ is finite and $f$ is injective, $f$ is also surjective. Then every $z \in G$ is of the form $x^{-1} \sigma(x)$ for some $x$.

Now let $z \in G$ with $z = x^{-1} \sigma(x)$. We have $\sigma(z) = \sigma(x^{-1} \sigma(x)) = \sigma(x)^{-1} x = (x^{-1} \sigma(x))^{-1} = z^{-1}$. Thus $\sigma$ is in fact the inversion mapping, and we assumed that $\sigma$ is a homomorphism. By a previous example, then, $G$ is abelian.

• steve  On January 9, 2012 at 6:14 am

Hi, firstly thanks for the proof. I would never imagined that the proof actually requires defining a new function. Could you please explain a little bit about the last line? “By a previous example, then, G is abelian.” I don’t see how G is abelian…

• steve  On January 9, 2012 at 6:15 am

Nevermind, got it =D

• nbloomf  On January 9, 2012 at 1:45 pm

In case anyone else is confused, if the inversion map is a homomorphism then we have $y^{-1}x^{-1} = (xy)^{-1} = x^{-1}y^{-1}$ for all $x,y \in G$. Since every element is an inverse, then, $xy = yx$ for all $x,y \in G$.