# Necessary and sufficient condition for isomorphism

• Jul 2nd 2011, 09:26 AM
Jagger
Necessary and sufficient condition for isomorphism
Hi everyone!

I have the next problem:

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

Any suggestion?

• Jul 2nd 2011, 10:32 AM
Isomorphism
Re: Necessary and sufficient condition for isomorphism
Quote:

Originally Posted by Jagger
Hi everyone!

I have the next problem:

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

Any suggestion?

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

$\varphi^p$ will stand for $\varphi$ composed with itself 'p' times. (Naturally $p \in \mathbb{N}$)
Exercise 1: Prove that if $\varphi$ is an isomorphism then $\varphi^p$ is an isomorphism too.

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

Step 1:
Let $(m,n)=d$. Then write ma + nb = d for some integers a and b (Why can you do this?).
$\varphi^{a}(g) = g^{ma} = g^d$
[Can I do the above step if 'a' is negative?]

Exercise 2:What happens, if $d > 1$?
[Hint: Look at $\text{ker}(\varphi^a)$ and rule out this case]

If d=1, we have $\varphi^a(g) = g \implies \varphi^{2a}(g) = g^2$ and,
Exercise 3: $\varphi^{2a}$ is a homomorphism iff G is abelian (Prove it!).

Step 2:
Exercise 4:Verify: If G is abelian and $(m,n)=1$, then $\varphi$ is an isomorphism.
This part is easy. Use the abelian nature to prove the function is a homomorphism and use $ma + nb = 1$ to prove $\text{ker}(\varphi) = {e}$.
__________________________________________________ _____________________________

Done! :)
• Jul 3rd 2011, 12:22 PM
Jagger
Re: Necessary and sufficient condition for isomorphism
Thanks for your answer! I forgot to said that $G$ was abelian but you know this. I don't understand in the exercise 4 how using that $ma+nb=1$ you can prove $ker(\varphi)=e$ and if you prove this you have that $\varphi$ is an isomorphism? or you have to prove that $\varphi$ is surjective too?

Thanks again!
• Jul 3rd 2011, 01:46 PM
Isomorphism
Re: Necessary and sufficient condition for isomorphism
Quote:

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

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

Do we need to prove surjectivity in this case? Notice that the map $\varphi: G \to G$. In this case surjectivity is equivalent to injectivity.
• Jul 3rd 2011, 04:15 PM
Deveno
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 $g^m = e$.

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.....