Fixed point free automorphism of order 2

[PragmaticYak]

Homework Statement

(Problem 1.6.23 from Dummit and Foote, 3rd edition)

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

Relevant Equations

If φ is a group homomorphism, then φ(x^{-1}) = φ^{-1}(x).

I did not use the hint for this problem. Here is my attempt at a proof:

Proof: Note first that $σ(σ(x)) = x$ for all $x \in G$. Then $σ^{-1}(σ(σ(x))) = σ(x) = σ^{-1}(x) = σ(x^{-1})$.

Now consider $σ(gh)$ for $g, h \in G$. We have that $σ(gh) = σ((gh)^{-1}) = σ(h^{-1}g^{-1})$. Additionally, $σ(gh) = σ(g)σ(h) = σ(g^{-1})σ(h^{-1}) = σ(g^{-1}h^{-1})$. Thus we have $σ(h^{-1}g^{-1}) = σ(gh) = σ(g^{-1}h^{-1}$). Since $σ$ is a bijection, $h^{-1}g^{-1} = g^{-1}h^{-1}$, which implies $gh = hg$. Thus $G$ is abelian.

The thing about this proof that troubles me is that it seems to imply that for all elements $x \in G$, $x = x^{-1}$. Is this really a necessary consequence of having such an automorphism on $G$? Or is there a mistake in my proof? The only part of the proof I can think of that might be problematic is where I apply $σ^{-1}$ to the composition, but it seems perfectly valid because $σ^{-1}$ exists. Or is there a problem somewhere else? Thank you in advance.

Edited to add LaTeX formatting.

 

[andrewkirk]

Your last equals sign is wrong:
$$\sigma^{-1}(x) = \sigma(x^{-1})$$
I think what you meant is the property we get from automorphism being a homomorphism, which is that:
$$(\sigma(x))^{-1}= \sigma(x^{-1})$$
The second is correct. The first is not.
Note that in the second one an exponent of -1 always means "take the inverse in this group", whereas in the first one, the first -1 exponent means "take the inverse function", which is a completely different thing: nothing to do with group inverses.

 

[PragmaticYak]

Took another crack at this problem. I’ll show the part after proving that every element of $G$ can be written as $x^{-1}\sigma(x)$.

Let $x \in G$. Then $x = y^{-1}\sigma(y)$ for some $y \in G$. Now $\sigma(x) = \sigma(y^{-1}\sigma(y)) = \sigma(y^{-1})y$. Rearranging, $y^{-1} = \sigma(x)^{-1}\sigma(y^{-1}) = \sigma(x^{-1}y^{-1})$. Plugging this back into the original equation, we have $x = \sigma(x^{-1}y^{-1})\sigma(y) = \sigma(x^{-1}y^{-1}y) = \sigma(x^{-1})$.

Now let $a, b \in G$. Then $ab = \sigma(a^{-1})\sigma(b^{-1}) = \sigma(a^{-1}b^{-1})$. Additionally, $ab = \sigma((ab)^{-1}) = \sigma(b^{-1}a^{-1})$. Since $\sigma$ is injective, $a^{-1}b^{-1} = b^{-1}a^{-1}$, meaning $ab = ba$. Thus $G$ is abelian.