# Fixed point free automorphism of order 2

• PragmaticYak

#### 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.

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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.

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.