If a group has an automorphism which is fixed point free of order 2, then it is abelian

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


We define a mapping f : G \rightarrow G by f(x) = x^{-1} \sigma(x).

Claim: f is injective.
Proof of claim: Suppose f(x) = f(y). Then y^{-1} \sigma(y) = x^{-1} \sigma(x), so that xy^{-1} = \sigma(x)\sigma(y^{-1}), and xy^{-1} = \sigma(xy^{-1}). Then we have xy^{-1} = 1, hence x = y. So f is injective.

Since G is finite and f is injective, f is also surjective. Then every z \in G is of the form x^{-1} \sigma(x) for some x.

Now let z \in G with z = x^{-1} \sigma(x). We have \sigma(z) = \sigma(x^{-1} \sigma(x)) = \sigma(x)^{-1} x = (x^{-1} \sigma(x))^{-1} = z^{-1}. Thus \sigma is in fact the inversion mapping, and we assumed that \sigma is a homomorphism. By a previous example, then, G is abelian.

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Comments

  • steve  On January 9, 2012 at 6:14 am

    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.

    • steve  On January 9, 2012 at 6:15 am

      Nevermind, got it =D

      • nbloomf  On January 9, 2012 at 1:45 pm

        In case anyone else is confused, if the inversion map is a homomorphism then we have y^{-1}x^{-1} = (xy)^{-1} = x^{-1}y^{-1} for all x,y \in G. Since every element is an inverse, then, xy = yx for all x,y \in G.

Trackbacks

  • By Group isomorphisms on February 11, 2014 at 11:00 am

    […] #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… […]

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