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Let $G_k$ and $G_m$ be cyclic groups of orders $k,m$ respectively. Is there a way to count the number of epimorphisms from $G_m$ to (on) $G_k$?

Thank you.

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    Could you please elaborate or explain why you believe it would be the number of divisors of $m$ that are coprime with $k$?2012-12-24
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    Or on second thought, that's incorrect... the observation is that if b generates $G_m$, then for $f$ to be an epimorphism it must fulfill $f(b)=a^q$, where $q$ is coprime with $k$ (and this determines the morphism entirely). For such an epimorphism to exist it must be that $o(a^q)=k$ divides $o(b)=m$.2012-12-24
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    And then I think I've been too hasty altogether.2012-12-24
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    Yeah, ok. So $k$ must divide $m$, and given that there are $t(k)$ unique ways to set $f(b)$ where $t$ is the totient function, hence $t(k)$ epimorphisms. At least that looks to be the implication. What do you guys think? Am I way off?2012-12-24
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    Here's the argument, more coherently. We notice that if $C_n$ is cyclic of order n generated by $a$ and G is a group, then there's a homomorphism from $C_n$ to G such that $f(a)=g$ iff $o(g)|o(a)=n$. Let $f$ be an epimorphism from $C_m$ to $C_k$, and let $b$ generate $C_m$. Then $f(b)=a^q$ for some q and this determines the epimorphism. But since it's an epimorphism, $a^q$ must generate $C_k$, therefore $(q,k)=1$. Furthermore, by the above, f is a homomorphism iff $o(a^q)=k|o(b)=m$.2012-12-24

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Take $G_n=\mathbb Z_n$. In order to have an epimorphism $f:\mathbb Z_m\to\mathbb Z_k$ we must have $k\mid m$ ($\overline 1\in\operatorname{Im}(f)$ $\Leftrightarrow$ $\exists\hat{a}\in\mathbb Z_m$ such that $f(\hat{a})=\overline 1$ $\Rightarrow$ $mf(\hat{a})=\overline m$ $\Rightarrow$ $\overline 0=\overline m$ $\Rightarrow$ $k\mid m$). Furthermore, every homomorphism will be determined by its value on $\hat 1$. Since we want the image to cover all $\mathbb Z_k$, then the image of $\hat 1$ must be invertible (as an element of the ring $\mathbb Z_k$). Conclusion: we have $\phi(k)$ epimorphisms, where $\phi$ is the Euler totient function.