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.
Best Answer
Set $k=|g|$ for convenience. Then $\phi(g)^k=\phi(g^k)=\phi(e)=e$. Hence the order of $\phi(g)$ divides $k$.