Hi everyone!

I have the next problem:

a finit group which has order and defined by , I need to find a necessary and sufficient condition so that is an isomorphism.

Any suggestion?

Thanks in advance!

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- Jul 2nd 2011, 08:26 AMJaggerNecessary and sufficient condition for isomorphism
Hi everyone!

I have the next problem:

a finit group which has order and defined by , I need to find a necessary and sufficient condition so that is an isomorphism.

Any suggestion?

Thanks in advance! - Jul 2nd 2011, 09:32 AMIsomorphismRe: Necessary and sufficient condition for isomorphism
__Claim:__

defined by is an isomorphism iff G is abelian and

will stand for composed with itself 'p' times. (Naturally )

**Exercise 1:**Prove that if is an isomorphism then is an isomorphism too.

Assume is an isomorphism and you can prove the conditions in two steps:

__Step 1:__

Let . Then write ma + nb = d for some integers a and b (Why can you do this?).

[Can I do the above step if 'a' is negative?]

**Exercise 2:**What happens, if ?

[Hint: Look at and rule out this case]

If d=1, we have and,

**Exercise 3:**is a homomorphism iff G is abelian (Prove it!).

__Step 2:__

**Exercise 4:**Verify: If G is abelian and , then is an isomorphism.

This part is easy. Use the abelian nature to prove the function is a homomorphism and use to prove .

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Done! :) - Jul 3rd 2011, 11:22 AMJaggerRe: Necessary and sufficient condition for isomorphism
Thanks for your answer! I forgot to said that was abelian but you know this. I don't understand in the exercise 4 how using that you can prove and if you prove this you have that is an isomorphism? or you have to prove that is surjective too?

Thanks again! - Jul 3rd 2011, 12:46 PMIsomorphismRe: Necessary and sufficient condition for isomorphism
- Jul 3rd 2011, 03:15 PMDevenoRe: Necessary and sufficient condition for isomorphism
for maps on finite sets, injective implies surjective and vice versa.

now think about what it means for g to be in ker(φ). it means .

there are two ways this could happen: g = e, or |g| divides m. suppose that |g| divides m.

since |g| also divides |G| = n, |g| is a common divisor, so.....