Results 1 to 2 of 2

Thread: orders of elements and isomorphisms

  1. #1
    Member
    Joined
    Aug 2010
    Posts
    77

    orders of elements and isomorphisms

    Show that if $\displaystyle \gamma: G_1 \rightarrow G_2$ is an isomorphism, then for each $\displaystyle g \in G_1, o(\gamma(g)) = o(g)$

    Does this argument work?

    if $\displaystyle g \in G_1 $ has order l $\displaystyle \Rightarrow g^l = e_1$ then
    $\displaystyle \gamma (g^l) = \gamma (e_1)$
    then as the identity element of $\displaystyle G_1$ is mapped to the identity element $\displaystyle e_2 \in G_2$
    $\displaystyle \gamma (g^l) = \gamma (e_1) = e_2$
    then as $\displaystyle \gamma(x_1x_2)= \gamma(x_1)\gamma(x_2)$ must hold
    we have
    $\displaystyle \gamma(g^l) = \gamma(g) ^l = (g')^l = e_2$ where as bijection each $\displaystyle g' = \gamma(g)$
    therefore $\displaystyle (g')^l = e_2 $ in $\displaystyle \in G_2$
    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 FGT12 View Post
    Show that if $\displaystyle \gamma: G_1 \rightarrow G_2$ is an isomorphism, then for each $\displaystyle g \in G_1, o(\gamma(g)) = o(g)$

    Does this argument work?

    if $\displaystyle g \in G_1 $ has order l $\displaystyle \Rightarrow g^l = e_1$ then
    $\displaystyle \gamma (g^l) = \gamma (e_1)$
    then as the identity element of $\displaystyle G_1$ is mapped to the identity element $\displaystyle e_2 \in G_2$
    $\displaystyle \gamma (g^l) = \gamma (e_1) = e_2$
    then as $\displaystyle \gamma(x_1x_2)= \gamma(x_1)\gamma(x_2)$ must hold
    we have
    $\displaystyle \gamma(g^l) = \gamma(g) ^l = (g')^l = e_2$ where as bijection each $\displaystyle g' = \gamma(g)$
    therefore $\displaystyle (g')^l = e_2 $ in $\displaystyle \in G_2$
    This is only half the battle. You've shown that $\displaystyle \left|\gamma(g)\right|\mid\ell$, but you can use the same process in reverse (since all you used was that $\displaystyle \gamma$ is a homomorphism, but $\displaystyle \gamma^{-1}$ is as well!) to get that $\displaystyle \ell\mid\left|\gamma(g)\right|$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Orders of elements of Dihedral Group D8
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Oct 12th 2011, 02:32 AM
  2. Orders of elements in Symmetric groups
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: Mar 30th 2010, 02:39 PM
  3. About orders of elements(Algebraic structures-subgroups)
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: Jan 12th 2010, 07:24 AM
  4. Orders of elements
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Jan 9th 2009, 11:15 AM
  5. Replies: 1
    Last Post: Oct 8th 2007, 05:13 PM

Search Tags


/mathhelpforum @mathhelpforum