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**Amer** let G be a finite group

if sigma is automorphism from G to G such that

$\displaystyle \sigma (g) = g \text{ if and only if }\;\; g =1 $

given that $\displaystyle \sigma^2 $ is the identity automorphism

prove that G is abelian ( hint prove that every element of G can be written as $\displaystyle x^{-1}\sigma (x) $ apply sigma to such expression

$\displaystyle x = x^{-1} \sigma (x) $

$\displaystyle \sigma (x) = \sigma (x^{-1} \sigma (x)) $

$\displaystyle \sigma (x) = \sigma (x^{-1}) \sigma (\sigma (x)) $

$\displaystyle \sigma (x) \sigma (x) = \sigma (x) \sigma (x^{-1}) \sigma (\sigma (x)) $

$\displaystyle ( \sigma (x) )^2 = \sigma (\sigma (x)) $

as you can see I end up with what I am asking about, did I made something wrong?

I work in it in another way

$\displaystyle \sigma ^2 (x) = x $ since sigma^2 is the identity

let $\displaystyle \sigma (x) = x* $

$\displaystyle \sigma (\sigma (x)) = \sigma (x*) \Rightarrow \;\; x = \sigma (x*) $

so if x image is x* then x* image is x