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This question coutesy of Allan Clark's "Elements of Abstract Algebra" (60$\zeta$).

Find the number of homomorphisms from $\Bbb{Z}_m\to \Bbb{Z}_n$ as a function of $m$ and $n$.

This is stumping me, can anyone help?

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In general, if $G$ is any group, then you can show that the number of homomorphisms $\mathbb{Z}_m \rightarrow G$ is the number of solutions to $x^m = 1$ in $G$. How many solutions are there to $x^m = 1$ in $\mathbb{Z}_n$?

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  • $\begingroup$ Also we can say that number of homomorphism $\mathbb Z_m \rightarrow G$ is the number of solution to mx = 0 in G $\endgroup$ – user120386 Mar 5 '14 at 16:08
  • $\begingroup$ Do you have a proof? This is interested to me. $\endgroup$ – Mr.Lilly Jan 31 '15 at 16:26
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HINT: If $h:\Bbb Z_m\to\Bbb Z_n$ is a homomorphism, $\big|h[\Bbb Z_m]\big|\cdot|\ker h|=m$, and $\big|h[\Bbb Z_m]\big|$ divides $n$.

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I am not sure if this is the correct way to go but look good to me hence, i am sharing.

Let $f$ be the homomorphism from $Z_{m} \to Z_{n}$

Since $O(f([1]_{n})) \space | \space O(Z_{n}) \implies O(f([1]_{n})) \space | \space n$

Also $O(f([1]_{n})) \space | \space m $

Hence number of homomorphism = $gcd(n, m)$

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