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

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

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

Quote:

Originally Posted by

**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

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

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

Quote:

I think you are making the mistake of interpreting $\displaystyle (\sigma^2)(x)=(\sigma(x))^2$ and not $\displaystyle \sigma(\sigma(x))$, 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 $\displaystyle \sigma $ such that $\displaystyle \sigma (g) = g$ if and only if g=1 .

if $\displaystyle \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 $\displaystyle x^{-1}\sigma (x) $ and apply $\displaystyle \sigma $ to such expression ]

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

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

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

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

Quote:

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

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

so I suppose that

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

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

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

then I multiply both sides by sigm(x)

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

$\displaystyle ( \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

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

Quote:

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

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

so I suppose that

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

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

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

Quote:

Originally Posted by

**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

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

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

Swlabr, NonCommAlg Thanks very much it is very clear now