orders of elements and isomorphisms

**FGT12** · December 31st 2010, 08:50 AM

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 $\in G_2$

**Drexel28** · December 31st 2010, 10:38 AM

Originally Posted by FGT12

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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