Let $\phi : G \to H $be a group homomorphism. Use the First Isomorphism Theorem to show that if $g \in G$ has finite order, then ord($\phi(g))|$ord$(g)$.
My problem here is having to use the first isomorphism theorem, as I can prove this without. So far I have considered the order of $g$. Suppose $g^n = e_G$. Then (in the following $\overline{\phi}$ is the induced homomorphism $\overline{\phi}: G/(N=ker \phi) \to Im(\phi) \in H$. Then:
$(\overline{\phi}(gN))^n = (\phi(g))^n$ $ = \phi(g^n)$ $= \phi (e_G)$ $= e_H$
I think I should be able to use this to show that the number of cosets of $N$ divides $n$ but I'm really not sure to go with this. Thanks for any help.
If you are being asked to use the First Isomorphism Theorem what seems natural is to prove that the order of $gN$ in $G/N$ divides the order of $g$ in $G$ for the normal subgroup $N = \ker \phi$ (and you have the basic argument for this written already it just needs some tweaking) and then use that $\phi(G) \cong G/N$ along with the fact that an isomorphism preserves the order of any element it is applied to in order to finish the proof.