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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.

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Finiteness is necessary as swapping the generators of the free group with two generators shows. –  Hagen von Eitzen Feb 13 '13 at 15:53
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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

2 Answers 2

up vote 4 down vote accepted

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.

Now come to your probelm:

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.

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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).

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+1 nice Alexander, as usual. –  B. S. Mar 9 '13 at 6:00

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