let G be a finite group
if sigma is automorphism from G to G such that
given that is the identity automorphism
prove that G is abelian ( hint prove that every element of G can be written as apply sigma to such expression
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
since sigma^2 is the identity
let
so if x image is x* then x* image is x
Thanks for your respondI think you are making the mistake of interpreting and not , no?
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 such that if and only if g=1 .
if 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 and apply to such expression ]
Page 38, Question 23. TextBook D.S.Dummit and R.M Foote, Abstract Algebra, edition, prentice-Hall, 2004. QS162.D85
so what I did wrong in thisthe multiplication in Aut(G) is functional composition, not the "element-wise" multiplication of G.
the question gave me a hint which said show that every element in G can be written as
so I suppose that
then i apply sigma to both sides
since sigma is homo
then I multiply both sides by sigm(x)
I cant see a mistake :S I know that the multiplication in Auto G is composition of functions
Thanks