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I am studying for a final and am trying to solve this problem: Let $G$ be a finite group with an automorphism $\sigma:G\rightarrow G$ such that $\sigma \circ \sigma=1$ and whose only fixed point is the identity element. I want to prove that $G$ is abelian and has odd order.

To show $G$ is abelian, since $\sigma \circ \sigma(xy)=xy$, showing that $xy=yx$ is the same as showing $\sigma \circ \sigma(xy)=\sigma \circ \sigma(yx)$. Since $\sigma \circ \sigma(xy)=\sigma(\sigma(x)\sigma(y))$, we want to show $\sigma(\sigma(x)\sigma(y))=\sigma(\sigma(y)\sigma(x))$, so it is enough to show that $\sigma(x)\sigma(y)=\sigma(y)\sigma(x)$, but I am having trouble working out the details.

To prove that $G$ has odd order, I am trying to show this by contradiction and assuming that $G$ is even. If $G$ is even, then it has at least one element $y$ of order 2, but using this, I haven't been able to get a contradiction.

dustin
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user196559
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3 Answers3

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The map $x^{-1}\sigma(x)$ is injective, proof: assume $x^{-1}\sigma(x)=y^{-1}\sigma(y)$ then $yx^{-1}=\sigma(y)\sigma(x)^{-1}=\sigma(yx^{-1})\implies yx^{-1}=e\implies y=x$ since there are no fixed points besides $e$. Since $G$ is finite and $x^{-1}\sigma(x)$ is injective it is also surjective.

hence we can write every element of $G$ in the form $x^{-1}\sigma(x)$. Take $g$ in $G$, it can be written as $x^{-1}\sigma(x)$. thus $\sigma(g)=\sigma(x)^{-1}x$. Notice multiplying $g$ with $\sigma(g)$ gives us $e$. we conclude $\sigma(g)=g^{-1}$

Two things are now clear:

  • $|G|$ cannot be even for if it were, it would have an element of order $2$, which would map to itself under $\sigma$, for it is its own inverse.
  • $G$ is abelian, for $\sigma(gh)=g^{-1}h^{-1}$ since $\sigma$ is homomorphism, but $\sigma(gh)=(gh)^{-1}=h^{-1}g^{-1}$. thus $g^{-1}h^{-1}=h^{-1}g^{-1}\implies hg=gh$ (just take inverses on both sides).
hbghlyj
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Asinomás
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For the second part, note that the elements $x\in G$ with $x\ne e$ can be paired off $\{x, \sigma(x)\}$. If there are $k$ such pairs, then there are $2k+1$ elements in $G$.

In general, given an involution $\sigma$ on a finite group $G$, the parity of $|G|$ is the same as the parity of the number of fixed points of $\sigma$.

lhf
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    Nice, this works for any involution on a finite set with an odd number of fixed points. – Asinomás Dec 16 '14 at 02:43
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    It can be seen also in the following way, write the permutation in cycle decomposition (writing the trivial cycles of length 1 for some reason). Since it is an involution all cycles have length $1$ or $2$, hence the parity of the group is determined by the number of $1$-cycles or fixed points. – Asinomás Dec 16 '14 at 02:48
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Note that $ \sigma \circ \sigma =\iota,$ the identity automorphism is given. Thus, for all $g\in G,$ we have $$g=\sigma ^2 (g)=\sigma(\sigma(g)). $$Since $\sigma $ is an automorphism , $\sigma ^{-1}$ exists and is a homomorphism. Therefore, $\sigma ^{-1}(g)=\sigma^{-1}(\sigma(\sigma(g)))=\sigma(g).$ Because $\sigma$ is a homomorphism, an elementary exercise shows that $\sigma ^{-1}(g)=\sigma (g^{-1}).$ We then have $\sigma (g^{-1}h^{-1})=\sigma(g^{-1})\sigma(h^{-1})=\sigma (g)\sigma(h)=\sigma((hg)^{-1})=\sigma(hg).$ But this yields the equality $\sigma(hg)=\sigma(gh)$ which in turn yields $hg=gh$ upon applying $\sigma ^{-1}$ both sides. Since the only fixed point of the automorphism is $1,$ for non trivial $g, g$ and $\sigma (g)$ are different so $$G=\cup _{ g \in G\setminus {1}}\{g, \sigma(g)\} \cup \{1\}$$ giving $|G|=2|G\setminus \{1\}|+1$ which is odd .

Tendai Shumba
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