Results 1 to 6 of 6

Math Help - Prove isomorphism

  1. #1
    Member
    Joined
    Feb 2010
    Posts
    84

    Prove isomorphism

    If G and H are groups and f:G --> H is an isomorphism, prove that a and f(a) have the same order, for any a E G

    I understand the concept. Isomorphism has one to one property, hence a have the same order of f(a) ... however, I don't not know how to prove it mathematically.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by rainyice View Post
    If G and H are groups and f:G --> H is an isomorphism, prove that a and f(a) have the same order, for any a E G

    I understand the concept. Isomorphism has one to one property, hence a have the same order of f(a) ... however, I don't not know how to prove it mathematically.
    Notice that if |g|=n then \theta(g)^n=\theta(g^n)=\theta(e_G)=e_H and so from basic group theory |\theta(g)|\mid n. But! Using the exact same idea and noting that \theta^{-1}:H\to G is an isomorphism we note that n\mid|\theta(g)| and so |theta(g)|=\pm n but since |\theta(g)|\geqslant 0 it easily follows that |\theta(g)|=n
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2010
    Posts
    84
    Quote Originally Posted by Drexel28 View Post
    Notice that if |g|=n then \theta(g)^n=\theta(g^n)=\theta(e_G)=e_H and so from basic group theory |\theta(g)|\mid n. But! Using the exact same idea and noting that \theta^{-1}:H\to G is an isomorphism we note that n\mid|\theta(g)| and so |theta(g)|=\pm n but since |\theta(g)|\geqslant 0 it easily follows that |\theta(g)|=n
    I don't get where the |theta(g)| \ n comes from ..... ><"
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by rainyice View Post
    I don't get where the |theta(g)| \ n comes from ..... ><"
    If g^n=e then |g|\mid n
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Apr 2010
    Posts
    37
    Quote Originally Posted by Drexel28 View Post
    If g^n=e then |g|\mid n
    That's if we assume that group G is finite. However, can we assume that these two groups are finite?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by MissMousey View Post
    That's if we assume that group G is finite. However, can we assume that these two groups are finite?
    Assume that |\theta(g)|=\infty but |g|=n<\infty then \theta(g)^n=\theta(g^n)=\theta(e_G)=e_H and so |\theta(g)|\leqslant n<\infty, contradiction. You can do the reverse just as easily.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove c is an isomorphism iff G is abelian
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: November 1st 2010, 07:39 PM
  2. Prove that Phi^-1:W->V is an isomorphism
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 22nd 2010, 08:30 AM
  3. Prove isomorphism
    Posted in the Advanced Algebra Forum
    Replies: 20
    Last Post: May 18th 2010, 01:55 AM
  4. Prove isomorphism
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 6th 2010, 06:21 PM
  5. Replies: 4
    Last Post: February 14th 2010, 04:05 AM

Search Tags


/mathhelpforum @mathhelpforum