is f(f(x)) = f(x) f(x) , f is automorphism f^2 is identity

**Amer** · November 14th 2011, 09:11 AM

let G be a finite group
if sigma is automorphism from G to G such that

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

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

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

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

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

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

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

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

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

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

let $\sigma (x) = x*$

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

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

**Drexel28** · November 14th 2011, 11:00 AM

Originally Posted by Amer

let G be a finite group
if sigma is automorphism from G to G such that

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

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

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

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

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

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

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

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

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

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

let $\sigma (x) = x*$

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

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

I think you are making the mistake of interpreting $(\sigma^2)(x)=(\sigma(x))^2$ and not $\sigma(\sigma(x))$ , no?

**Amer** · November 14th 2011, 07:06 PM

I think you are making the mistake of interpreting $(\sigma^2)(x)=(\sigma(x))^2$ and not $\sigma(\sigma(x))$ , no?

Thanks for your respond
I think it is wrong but I reached it from the given hint tell me if I did something wrong in my work

I want to give the whole question as it is in the book

Let G be a finite group which possesses an automorphism $\sigma$ 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 called 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 expression ]

Page 38, Question 23. TextBook D.S.Dummit and R.M Foote, Abstract Algebra, $3^{rd}$ edition, prentice-Hall, 2004. QS162.D85

**Deveno** · November 14th 2011, 08:09 PM

the multiplication in Aut(G) is functional composition, not the "element-wise" multiplication of G.

**Amer** · November 15th 2011, 01:27 AM

the multiplication in Aut(G) is functional composition, not the "element-wise" multiplication of G.

so what I did wrong in this
the question gave me a hint which said show that every element in G can be written as

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

so I suppose that

$x = x^{-1} \sigma (x)$ then i apply sigma to both sides

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

$\sigma (x) = \sigma (x^{-1}) \sigma (\sigma (x))$ since sigma is homo
then I multiply both sides by sigm(x)

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

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

I cant see a mistake :S I know that the multiplication in Auto G is composition of functions
Thanks

**Swlabr** · November 15th 2011, 01:36 AM

Originally Posted by Amer

so what I did wrong in this
the question gave me a hint which said show that every element in G can be written as

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

so I suppose that

$x = x^{-1} \sigma (x)$ then i apply sigma to both sides

No - every element can be written as $x^{-1}\sigma(x)$ , but the $x$ may vary. Basically, for every $y\in G$ there exists an $x\in G$ such that $y=x^{-1}\sigma(x)$ .

**NonCommAlg** · November 15th 2011, 03:04 AM

Originally Posted by Amer

let G be a finite group
if sigma is automorphism from G to G such that

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

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

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

define the map $f:G \longrightarrow G$ by $f(x) = x^{-1}\sigma(x)$ for all $x \in G.$ then $f$ is one-to-one because if $f(x)=f(y),$ then $\sigma(xy^{-1})=xy^{-1}$ and so $x= y,$ by the first condition you gave us. thus $f$ is onto because $G$ is finite. now let $g \in G$ and choose $x \in G$ such that $g=f(x)=x^{-1}\sigma(x).$ then $\sigma(g)=\sigma(x^{-1})x = g^{-1}.$ hence $\sigma(g)=g^{-1}$ for all $g \in G$ and so $G$ is abelian.

**Amer** · November 15th 2011, 08:59 AM

Swlabr, NonCommAlg Thanks very much it is very clear now

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