# Thread: Necessary and sufficient condition for isomorphism

1. ## 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?

2. ## Re: Necessary and sufficient condition for isomorphism

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?

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!

3. ## 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!

4. ## Re: Necessary and sufficient condition for isomorphism

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.

5. ## 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.....

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