# The existence of a group automorphism with some properties implies commutativity.

Let $G$ be a finite group, $T$ be an automorphisom of $G$ st $Tx = x \iff x=e$. Suppose further that $T^2 =I$. Prove that $G$ is abelian.

I was thinking if I show $T aba^{-1} b^ {-1}=aba^ {-1}b^{-1} \forall a, b \in G$. But I was unable to show it. Please give me any hints about it.

• Finiteness is necessary as swapping the generators of the free group with two generators shows. – Hagen von Eitzen Feb 13 '13 at 15:53
• If the claim is true, then necessarily $Tx=x^{-1}$ because (knowing $G$ is abelian) $T(xTx)=TxT^2x=Txx=xTx$, hence $xTx=e$ – Hagen von Eitzen Feb 13 '13 at 15:56

Hint: $|G|<\infty$, $T$ is an automorphism with the given property then $\forall g\in G$ can be written is of the form $g=x^{-1}T(x)$ to prove this result just define $f(x)=x^{-1}T(x)$ and show that $f$ is onto.
By above result $\forall a\in G$ we can write $a=x^{-1}T(x)$ for some $x\in G$ so $T(a)=T(x^{-1})T^2(x)=T(x^{-1}) x=[x^{-1}T(x)]^{-1}=a^{-1}$ so $T(ab)=(ab)^{-1}=b^{-1}a^{-1}$ so $T(a)T(b)=b^{-1}a^{-1}$ so $a^{-1}b^{-1}=b^{-1}a^{-1}$ so $ab=ba$, so $G$ is abelian.
• You don't really needs that $f$ is an automorphism, just that it is onto, which is much easier to prove. – Thomas Andrews Feb 13 '13 at 16:17
HINT: Consider $\sigma:x\mapsto x^{-1}T(x)$.
Note that, in general, if $f(x)$ is a fixed point free automorphism of $G$ (fixed point free means $x=1\Leftrightarrow f(x)=x$), then $\sigma(x)=x^{-1}f(x)$ is a bijection from $G$ to $G$ (though this function is not in general a homomorphism).