Exercise 1.6.23 ( Fixed point free automorphism)

Question

Let $G$ be a finite group which possesses an automorphism $\sigma$ (cf. Exercise 20) such that $\sigma(g) = g$ if and only if $g = 1$. If $\sigma^2$ is the identity map from $G$ to $G$, prove that $G$ is abelian (such an automorphism $\sigma$ is called {\bf fixed point free} of order $2$). [Show that every element of $G$ can be written in the form $x^{-1} \sigma(x)$ and apply $\sigma$ to such an expression.]

Richard Ganaye · Accepted Answer

\begin{proof} 
By hypothesis, the map $\sigma : G 	o G$ satisfies 
\begin{enumerate}
\item[(i)] $\sigma \in \mathrm{Aut}(G)$,
\item[(ii)] $\sigma^2 = \mathrm{Id}_G$,
\item[(iii)] $\forall g \in G,\ (\sigma(g) = g \iff g = 1_G)$.
\end{enumerate}

Consider the map
$$
	au
\left\{
\begin{array}{ccl}
G & 	o & G\
x & \mapsto &x^{-1}\sigma(x)
\end{array}
ight.
$$
(Warning: Since we don't know that $G$ is abelian, we cannot prove that $	au$ is a homomorphism.)

We prove that $	au$ is injective. For all $x,\, y \in G$,
\begin{align*}
	au(x) = 	au(y) &\Rightarrow x^{-1} \sigma(x) = y^{-1} \sigma(y)\
&\Rightarrow \sigma(xy^{-1}) = xy^{-1}\
&\Rightarrow xy^{-1} = 1_G&	ext{(by (iii))}\
& \Rightarrow x = y.
\end{align*}
This proves that $	au : G 	o G$ is injective. Since $G$ is a finite set, $	au$ is also surjective. Therefore every element  $z \in G$ can be written in the form $z =x^{-1} \sigma(x)$ for some $x \in G$.

\bigskip
Let $z$ be any element of $G$. By the preceding part, $z = x^{-1} \sigma(x)$ for some $x \in G$. Therefore
\begin{align*}
\sigma(z) &=\sigma(x^{-1} \sigma(x))\
&= \sigma(x)^{-1} x &(	ext{by (ii))}\
&= (x^{-1} \sigma(x))^{-1}\
&= z^{-1},
\end{align*}
thus
\begin{align}
\forall z \in G,\ \sigma(z) = z^{-1}.
\end{align}
Since $\sigma:G 	o G$ is a homomorphism (by (i)), where $\sigma$ is the map $z \mapsto z^{-1}$, we know that $G$ is abelian (as in Exercise 17) : for all $g,h \in G$,
\begin{align*}
h g &= (h^{-1})^{-1} (g^{-1})^{-1}\
&= \sigma(h^{-1}) \sigma(g^{-1}) &(	ext{by (1))} \
&= \sigma(h^{-1} g^{-1})& 	ext{(since $\sigma$ is a homomorphism)}\
&= \sigma((gh)^{-1})&(	ext{by (1))}\
&= ((gh)^{-1})^{-1}\
&= gh.
\end{align*}
If there is some map $\sigma: G 	o G$ such that (i), (ii), (iii) are true, then $G$ is abelian.
\end{proof}

Exercise 1.6.23 ( Fixed point free automorphism)

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Exercise 1.6.23 ( Fixed point free automorphism)

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