# Prove the order of the group homomorphism of an element divides the order of the element.

Let $\phi : G \rightarrow H$ be a group homomorphism. Prove $\forall g \in G$, the order of $\phi(g)$ divides $g$.

I've gotten to the point where I've shown that if,

$ord(\phi(g)) < ord(g)$ then $ord(\phi(g))$ divides $ord(g)$.

But what I've done to just show this seems unnecessarily complicated.

Set $k=|g|$ for convenience. Then $\phi(g)^k=\phi(g^k)=\phi(e)=e$. Hence the order of $\phi(g)$ divides $k$.
Let $n$ be the order of $g$.
Then $1=\phi(1)=\phi(g^n)=\phi(g)^n$. This implies that $n$ is a multiple of the order of $\phi(g)$.