Showing that $o(\phi(g))|o(g)$ if $\phi:G_1\mapsto G_2$ is a homomorphism and $g\in G_1$

Question

$o(g)$ denotes the order of $g$.

This is how I think I proved it:

Let $m:=o(g)$

Let $d=o(\phi(g))$. $~d\le m$ because $\langle\phi (g)\rangle=\{\phi(g),\phi(g)^2=\phi(g^2),...,\phi(g^m)=e_{G_2}\}$

If $o(\phi(g))=d$ by euclidean division there exist $q\in\Bbb N$ and $0\le r<d$ such that $$m=dq+r~\text{ and }~\phi(g)^d=e=\phi(g^d)\\ e=g^m=g^{dq+r}=g^{dq}g^r \implies g^r=g^{-dq}\\ \phi(g)^r=\phi(g^r)=\phi(g^{-dq})=(\phi(g)^d)^{-q}=e$$ but $d>r$ is the order of $\phi(g)$ so $r$ must be zero.

I'm not sure if this is all correct

Answer 1

You can get this more quickly using the First Isomorphism Theorem. Recall that $o(g) = |\langle g \rangle|$. Now $o(\phi(g)) = |\phi( \langle g \rangle)|$. Taking $G = \langle g \rangle$, the First Isomorphism theorem tells us that $|G| = |Im(\phi)| \cdot |\ker(\phi)|$. Note that $|Im(\phi)| = o(\phi(g))$.

Answer 2

I would just write $\phi (g)^r=e$, with $r\lt d$. So $r=0$. (It's pretty clear, since $e=\phi(g)^m=\phi(g)^d$).

But it seems to be correct.

Actually, once $\phi (g)^m=e$ you're done (by the little argument you essentially gave). That is, $g^a=e\implies o(g)\mid a$, generally.