Results 1 to 5 of 5

Math Help - Necessary and sufficient condition for isomorphism

  1. #1
    Newbie
    Joined
    Mar 2010
    From
    Uruguay
    Posts
    12

    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?

    Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6

    Re: Necessary and sufficient condition for isomorphism

    Quote Originally Posted by Jagger View Post
    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?

    Thanks in advance!
    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!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2010
    From
    Uruguay
    Posts
    12

    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!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6

    Re: Necessary and sufficient condition for isomorphism

    Quote Originally Posted by Jagger View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    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.....
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: October 16th 2011, 08:20 AM
  2. Sufficient condition for Existence of continous retraction
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: April 23rd 2010, 10:23 PM
  3. Replies: 4
    Last Post: February 14th 2010, 04:05 AM
  4. Necessary and sufficient condition for linear dependence
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: January 22nd 2010, 09:43 AM
  5. Replies: 0
    Last Post: June 4th 2008, 03:39 PM

Search Tags


/mathhelpforum @mathhelpforum