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I am not sure how to do this, because when I tried, it showed that $n$ divides $|\phi(g)|$? Let me show you my work so far...

For homomorphism $\phi:G\to H$, show that if $g\in G$ and $|g|=n$, then $|\phi(g)|$ divides $n$.

Let $|g|=n$ and $|\phi(g)|=m$. Then $g^{n}=e_{G}$ and $(\phi(g))^{m}=e_{H}$. Then there exists integers $q,r$ such that $m=nq+r$.

Then $e_{H}=(\phi(g))^{m}=\phi(g^{m})=\phi(g^{nq+r})=\phi(g^{nq})\phi(g^{r})=\phi(e_{G}^{q})\phi(g^{r})=(\phi(g))^{r}$.

Since $m$ is the smallest integer such that $(\phi(g))^{m}=e_{H}$, $r=0$. Then $m=nq$. Then $n$ divides $m$...

Yes, I feel like I am doing this very wrong...

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    $\begingroup$ Here's a simpler argument: Since $|g|=n$, we have $g^n=e_G$ which implies $[\phi(g)]^n=\phi(g^n)=\phi(e_G)=e_H$ and by definition of order of an element, $|\phi(g)|$ divides $n$. $\endgroup$ Mar 17 '18 at 22:06
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When you write $m=nq+r$, $r<n$ not $m$, so you can have $m=r$, you can write $n=mq+r, r<m$ $\phi(g)^n=\phi(g^n)=e_H=(\phi(g))^{mq}(\phi(g))^r=(\phi(g))^r$ implies that $r=0$ since $m$ is the minimal non zero integer such that $(\phi(g))^m=0$.

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  • $\begingroup$ I see what you mean... I was close but forgot some important things which caused my proof to collapse... Thank you for clearing everything up! $\endgroup$
    – user482939
    Mar 17 '18 at 21:52

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