# Thread: orders of elements and isomorphisms

1. ## 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 $\displaystyle \in G_2$

2. Originally Posted by FGT12
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 $\displaystyle \ell\mid\left|\gamma(g)\right|$.