Counting Homomorphisms from $\mathbb{Z}_n$ to $\mathbb{Z}_m$

Question

My notes have a proof of the following claim:

Prove there are $\gcd(n,m)$ homomorphisms $\phi: \mathbb{Z}_n \rightarrow \mathbb{Z}_m$.

I was able to follow most of it but I got lost. I outline the proof below and highlight where I get lost:

First we note that $\phi$ is completely determined by the image of $1$. To see this note: $\phi(x) = \phi(1+1+1+1....+1) = x \phi(1)$ so that if $\phi(1) = a \in \mathbb{Z}_n$ then the task is to find all $a$'s such that $\phi(x) = ax$ is a homomorphism. Arguments (which I don't reproduce) show that $\text{ord}(\phi(1))|n$ and $\text{ord}(\phi(1))|m$, thus $\text{ord}(\phi(1))|\gcd(m,n)$. Now I get lost. My notes say:

Look at $\langle\frac{m}{\gcd(m,n)}\rangle$ in $\mathbb{Z}_m$. Observe $\phi(1) \in \langle\frac{m}{\gcd(m,n)}\rangle$.

Thus $\phi(1) = k \frac{m}{\gcd(m,n)}$ for any $1 \leq k < \gcd(m,n)$. Thus there are $\gcd(m,n)$ homomorphisms.

Why do I look at $\langle\frac{m}{\gcd(m,n)}\rangle$ in $\mathbb{Z}_m$?
How do I see that $\phi(1)$ must be in $\langle\frac{m}{\gcd(m,n)}\rangle$?

This didn't quite help. Martin Toma's proof uses Chinese Remainder, which I haven't learned yet. The Gallian's paper and solution manual uses a sum of Euler functions equals $\gcd(m,n)$ - something I don't know how to prove. — yoshi, May 31 '18 at 19:02

Learnmore · Answer 1 · 2018-06-01T02:33:08.910

3

Any homomorphism $\phi :\Bbb Z_m\to \Bbb Z_n$ is determined by $\phi([1])$.

Now if $\phi(1)=a\implies \phi(m)=ma\implies 0=\phi(0)=ma\implies o(a)| m$

Also $a\in \Bbb Z_n\implies o(a)|n$

Combining $o(a)\mid \gcd(m,n)$ .

EDIT:

Now for each divisor $k$ of $n$, $\Bbb Z_n$ has a unique subgroup of order $k$ which has $\phi(k)$ many generators.

So for $a$ we have $\phi(k)$ many choices.

Hence total number of choices for $a$ will be $\sum_{k|\gcd(m,n)} \phi(k)$

It is a known result that $\sum_{k|\gcd(m,n)} \phi(k)=\gcd(m,n)$ Proof:Relation between $\gcd$ and Euler's totient function .

edited Jun 01 '18 at 02:33

answered May 31 '18 at 17:17

Learnmore

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I follow everything but the last line. Why does it follow there are $\gcd(m,n)$ homomorphims? – yoshi May 31 '18 at 17:23
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Shouldn't it be the number of divisors of $\text{gcd}(m,n)$ homomorphisms? For instance $\varphi: \mathbb{Z}_6 \to \mathbb{Z}_9$ can send $1$ to either $1$ or $3$ but the gcd is $3$ – N8tron May 31 '18 at 17:24
@yoshi; since the homo is determined by $a$ so number of homo should depend on the order of a – Learnmore May 31 '18 at 17:26
What can be the possible number of choices for $o(a)$ that is $gcd(m,n)$ – Learnmore May 31 '18 at 17:27
Ah I was missing another target you need the order of $a$ in $\mathbb Z_n$ so the possible targets in $\mathbb{Z}_9$ are $1$,$3$,$6$ – N8tron May 31 '18 at 17:27
@N8tron; i cleared things up in the comments – Learnmore May 31 '18 at 17:28
I'm still lost. I understand $o(a) | \gcd(m,n)$. But this only implies $o(a)$ divides the gcd. How do you get equality? – yoshi May 31 '18 at 17:31
@yoshi; $o(a)|m,o(a)|n\implies o(a)=\gcd(m,n)$ sorry for that – Learnmore May 31 '18 at 17:34
Why is that true? If $m=12$ and $n=8$ and $o(a) = 2$ then $2|12$ and $2|8$ but $2 \neq \gcd(12,8)$ – yoshi May 31 '18 at 17:37
@yoshi; see the edits – Learnmore May 31 '18 at 17:44
Can you provide an argument for why $o(a)$ has $\gcd(m,n)$ choices? It's not obvious to me. – yoshi May 31 '18 at 17:50
i provided dat only in the answer – Learnmore May 31 '18 at 17:53
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Sorry, to clarify -- why does $o(a) |m$, $o(a)|n$ imply there are $\gcd(m,n)$ possibilities for $o(a)$? – yoshi May 31 '18 at 17:57

score 0 · Answer 2 · 2020-09-20T06:05:24.050

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It is easy to see that $\phi(1)|\rm{gcd}(m,n)$, by Lagrange twice (and the homomorphism property).

But, for each divisor $k$ of $\rm{gcd}(m,n)$, there are $\varphi(k)$ different elements of order $k$. Thus we have $\sum_{k|{\rm{gcd}(m,n)}}\varphi(k)=\rm{gcd}(m,n)$ homomorphisms (by a result proved by Gauß, among others).

edited Sep 20 '20 at 06:05

answered May 31 '18 at 18:35

I see that, I'm just lost at the parts I bolded. – yoshi May 31 '18 at 18:37

Counting Homomorphisms from $\mathbb{Z}_n$ to $\mathbb{Z}_m$

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