- #1

PragmaticYak

- 4

- 1

- 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.

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.

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