Order of $\phi(g)$ divides the order of $g$

Let $\phi: G \to H$ be a group homomorphism, and let $g$ be an element of $G$. Show that the order of $\phi(g)$ divides the order of $g$.

Let $n=|g|, \nu = |\phi(g)|$. Since $g^n = e$ and $\phi(g^n) = \phi(g)^n$, we have $\nu \le n$.

Enormous hint: If $\nu \not\mid n$, write $n=k \nu +r$, with $r < \nu$.

The details:

Then $e=\phi(g)^n = (\phi(g)^\nu)^k \phi(g)^r = \phi(g)^r$, which contradicts the definition of $\nu$. Hence $\nu \mid n$.

Hint: If $g^n = e_G$, then $e_H = \phi(g^n) = \phi(g)^n$.

let $o(g) = n$ for $g\in G$ $\implies$ $g^{n} = e$_G

Now $\phi(g^{n}) = \phi(g)^{n}$ using group homomorphism property

as $\phi(g^{n}) = e$_H $\implies \phi(g)^{n} = e$_H

$\implies o(\phi(g)) <= n$ and $o(\phi(g))$ divides n