Necessary and sufficient condition for isomorphism

Hi everyone!

I have the next problem:

$\displaystyle G$ a finit group which has order $\displaystyle n$ and $\displaystyle \varphi:G\rightarrow G$ defined by $\displaystyle \varphi(a)=a^m\;\forall\;a\in G$, I need to find a necessary and sufficient condition so that $\displaystyle \varphi$ is an isomorphism.

Any suggestion?

Thanks in advance!

Re: Necessary and sufficient condition for isomorphism

Quote:

Originally Posted by

**Jagger** Hi everyone!

I have the next problem:

$\displaystyle G$ a finit group which has order $\displaystyle n$ and $\displaystyle \varphi:G\rightarrow G$ defined by $\displaystyle \varphi(a)=a^m\;\forall\;a\in G$, I need to find a necessary and sufficient condition so that $\displaystyle \varphi$ is an isomorphism.

Any suggestion?

Thanks in advance!

__Claim:__

$\displaystyle \varphi:G\rightarrow G$ defined by $\displaystyle \varphi(a)=a^m\;\forall\;a\in G, m > 1$ is an isomorphism iff G is abelian and $\displaystyle (m,n)=1$

$\displaystyle \varphi^p$ will stand for $\displaystyle \varphi$ composed with itself 'p' times. (Naturally $\displaystyle p \in \mathbb{N}$)

**Exercise 1:** Prove that if $\displaystyle \varphi$ is an isomorphism then $\displaystyle \varphi^p$ is an isomorphism too.

Assume $\displaystyle \varphi$ is an isomorphism and you can prove the conditions in two steps:

__Step 1:__

Let $\displaystyle (m,n)=d$. Then write ma + nb = d for some integers a and b (Why can you do this?).

$\displaystyle \varphi^{a}(g) = g^{ma} = g^d$

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

**Exercise 2:**What happens, if $\displaystyle d > 1$?

[Hint: Look at $\displaystyle \text{ker}(\varphi^a)$ and rule out this case]

If d=1, we have $\displaystyle \varphi^a(g) = g \implies \varphi^{2a}(g) = g^2$ and,

**Exercise 3:** $\displaystyle \varphi^{2a}$ is a homomorphism iff G is abelian (Prove it!).

__Step 2:__

**Exercise 4:**Verify: If G is abelian and $\displaystyle (m,n)=1$, then $\displaystyle \varphi$ is an isomorphism.

This part is easy. Use the abelian nature to prove the function is a homomorphism and use $\displaystyle ma + nb = 1$ to prove $\displaystyle \text{ker}(\varphi) = {e}$.

__________________________________________________ _____________________________

Done! :)

Re: Necessary and sufficient condition for isomorphism

Thanks for your answer! I forgot to said that $\displaystyle G$ was abelian but you know this. I don't understand in the exercise 4 how using that $\displaystyle ma+nb=1$ you can prove $\displaystyle ker(\varphi)=e$ and if you prove this you have that $\displaystyle \varphi$ is an isomorphism? or you have to prove that $\displaystyle \varphi$ is surjective too?

Thanks again!

Re: Necessary and sufficient condition for isomorphism

Quote:

Originally Posted by

**Jagger** Thanks for your answer! I forgot to said that $\displaystyle G$ was abelian but you know this. I don't understand in the exercise 4 how using that $\displaystyle ma+nb=1$ you can prove $\displaystyle ker(\varphi)=e$ and if you prove this you have that $\displaystyle \varphi$ is an isomorphism? or you have to prove that $\displaystyle \varphi$ is surjective too?

Thanks again!

Hint: $\displaystyle g^m = g^{ma +nb}$

Do we need to prove surjectivity in this case? Notice that the map $\displaystyle \varphi: G \to G$. In this case surjectivity is equivalent to injectivity.

Re: 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 $\displaystyle g^m = e$.

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.....