4
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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 $\phi(g)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.