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**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 $\displaystyle \in G_2$