Let be a finite group which possesses an automorphism such that if and only if . If is the identity map on , prove that is abelian. Such an automorphism is called *fixed point free* of order 2. [Hint: Show that every element of can be written in the form and apply to such an expression.]

We define a mapping by .

Claim: is injective.

Proof of claim: Suppose . Then , so that , and . Then we have , hence . So is injective.

Since is finite and is injective, is also surjective. Then every is of the form for some .

Now let with . We have . Thus is in fact the inversion mapping, and we assumed that is a homomorphism. By a previous example, then, is abelian.

## Comments

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…

Thanks in advance.

Nevermind, got it =D

In case anyone else is confused, if the inversion map is a homomorphism then we have for all . Since every element is an inverse, then, for all .

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[…] #2 Today, 12:00 Hi, This is a "standard" exercise. Here is a link to a proof: If a group has an automorphism which is fixed point free of order 2, then it is abelian | Project Cr… […]