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Thread: Necessary and sufficient condition for isomorphism

  1. #1
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    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!
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  2. #2
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    Re: Necessary and sufficient condition for isomorphism

    Quote Originally Posted by Jagger View Post
    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!
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  3. #3
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    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!
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  4. #4
    Lord of certain Rings
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    Re: Necessary and sufficient condition for isomorphism

    Quote Originally Posted by Jagger View Post
    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.
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  5. #5
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    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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