# Fixed point free automorphism of order 2

• PragmaticYak
In summary, the proof shows that a group with an automorphism ##\sigma## satisfying ##\sigma(\sigma(x)) = x## for all ##x \in G## must be abelian. This is because every element in the group can be written as the product of its inverse and the image of an element under ##\sigma##, and the fact that ##\sigma## is a homomorphism implies that the elements commute. However, the proof also raises the question of whether this property implies that every element in the group is equal to its own inverse, which may not be true.
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.

Last edited:
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.

## What is a fixed point free automorphism of order 2?

A fixed point free automorphism of order 2 is an automorphism (a bijective homomorphism from a mathematical structure to itself) such that applying it twice yields the identity automorphism, and it has no fixed points, meaning it maps every element of the structure to a different element.

## In which mathematical structures can fixed point free automorphisms of order 2 occur?

Fixed point free automorphisms of order 2 can occur in various mathematical structures, including groups, rings, fields, and topological spaces. They are particularly studied in the context of group theory and algebraic topology.

## Why are fixed point free automorphisms of order 2 significant in group theory?

In group theory, fixed point free automorphisms of order 2 are significant because they provide insights into the structure and symmetries of groups. They are used to study properties like group actions, and they can lead to the classification of groups and the understanding of their substructures.

## Can you provide an example of a fixed point free automorphism of order 2?

An example of a fixed point free automorphism of order 2 is the map $$f: \mathbb{Z}_n \to \mathbb{Z}_n$$ defined by $$f(x) = -x$$ in the additive group of integers modulo $$n$$, $$\mathbb{Z}_n$$, where $$n$$ is even. This map has order 2 because $$f(f(x)) = x$$ and is fixed point free because $$f(x) = x$$ only if $$x = 0$$ or $$x = n/2$$, and these are distinct elements mapped to each other.

## How are fixed point free automorphisms of order 2 used in algebraic topology?

In algebraic topology, fixed point free automorphisms of order 2 are used to study the properties of topological spaces and their symmetries. They can be used to construct and analyze double covers, study homotopy and cohomology theories, and investigate fixed point theorems in various topological contexts.

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