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Math Help - orders of elements and isomorphisms

  1. #1
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    orders of elements and isomorphisms

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

    Does this argument work?

    if g \in G_1 has order l  \Rightarrow g^l = e_1 then
    \gamma (g^l) = \gamma (e_1)
    then as the identity element of G_1 is mapped to the identity element e_2 \in G_2
    \gamma (g^l) = \gamma (e_1) = e_2
    then as \gamma(x_1x_2)= \gamma(x_1)\gamma(x_2) must hold
    we have
    \gamma(g^l) = \gamma(g) ^l = (g')^l = e_2 where as bijection each g' = \gamma(g)
    therefore (g')^l = e_2 in  \in G_2
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by FGT12 View Post
    Show that if \gamma: G_1 \rightarrow G_2 is an isomorphism, then for each g \in G_1, o(\gamma(g)) = o(g)

    Does this argument work?

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