Results 1 to 6 of 6

Thread: 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
    22
    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 $\displaystyle |g|=n$ then $\displaystyle \theta(g)^n=\theta(g^n)=\theta(e_G)=e_H$ and so from basic group theory $\displaystyle |\theta(g)|\mid n$. But! Using the exact same idea and noting that $\displaystyle \theta^{-1}:H\to G$ is an isomorphism we note that $\displaystyle n\mid|\theta(g)|$ and so $\displaystyle |theta(g)|=\pm n$ but since $\displaystyle |\theta(g)|\geqslant 0$ it easily follows that $\displaystyle |\theta(g)|=n$
    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 $\displaystyle |g|=n$ then $\displaystyle \theta(g)^n=\theta(g^n)=\theta(e_G)=e_H$ and so from basic group theory $\displaystyle |\theta(g)|\mid n$. But! Using the exact same idea and noting that $\displaystyle \theta^{-1}:H\to G$ is an isomorphism we note that $\displaystyle n\mid|\theta(g)|$ and so $\displaystyle |theta(g)|=\pm n$ but since $\displaystyle |\theta(g)|\geqslant 0$ it easily follows that $\displaystyle |\theta(g)|=n$
    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
    22
    Quote Originally Posted by rainyice View Post
    I don't get where the |theta(g)| \ n comes from ..... ><"
    If $\displaystyle g^n=e$ then $\displaystyle |g|\mid n$
    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 $\displaystyle g^n=e$ then $\displaystyle |g|\mid n$
    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
    22
    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 $\displaystyle |\theta(g)|=\infty$ but $\displaystyle |g|=n<\infty$ then $\displaystyle \theta(g)^n=\theta(g^n)=\theta(e_G)=e_H$ and so $\displaystyle |\theta(g)|\leqslant n<\infty$, 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: Nov 1st 2010, 06:39 PM
  2. Prove that Phi^-1:W->V is an isomorphism
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Sep 22nd 2010, 07:30 AM
  3. Prove isomorphism
    Posted in the Advanced Algebra Forum
    Replies: 20
    Last Post: May 18th 2010, 12:55 AM
  4. Prove isomorphism
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Apr 6th 2010, 05:21 PM
  5. Replies: 4
    Last Post: Feb 14th 2010, 03:05 AM

Search Tags


/mathhelpforum @mathhelpforum