All homomorphisms from Z_n to Z_m

[ArcanaNoir]

Homework Statement

Describe all group homomorphisms from \mathbb{Z}_n to \mathbb{Z}_m.

Homework Equations

\mathbb{Z}_n = {[0],[1],\dots ,[n-1]} with addition.

A homomorphism is an operation preserving map, ie \phi (a\ast b)=\phi (a) \# \phi (b).

One especially important homomorphism property is that \phi (a^k) = \phi (a)^k.

We can describe each homomorphism entirely by its action on any element that generate the group.

The Attempt at a Solution

I am pretty sure there are \text{gcd}(n,m) homomorphisms from \mathbb{Z}_n to \mathbb{Z}_m.

Based on some examples I worked out, I believe the solution is:

let [a] be any element which generates \mathbb{Z}_n

\phi ([a]) = \frac{n}{\text{gcd}(n,m)}\cdot k [a] where 0\le k < \text{gcd}(m,n)

 

[I like Serena]

You are correct, but your "description" can use improvement and I believe it contains a mistake.

Let's simplify your generator a bit.
Instead of picking [a], it's easier to pick [1] as a generator.

Then candidates for the homomophisms are the ones defined by ϕ([1]).
The restriction is that ϕ(n[1]) = n ϕ([1]) = 0 mod m.
In other words m | n ϕ([1]).

Now let g = gcd(m,n), then there are numbers a and b such that $m=ga, n=gb, \gcd(a,b)=1$.

Then:
$$ga\ |\ gbϕ([1]) \\
a\ |\ bϕ([1]) \\
a\ |\ ϕ([1])$$
Do you see why?

So you're left with the question which and how many values of ϕ([1]) meet with the condition that $\frac m {\gcd(m,n)}$ divides ϕ([1]).

 

[ArcanaNoir]

I did have [1] in mind for [a], but it should work on any generator.

I'm not sure what you are getting at with your last statement.

You mentioned you thought there was a mistake, but you also said I was correct?

 

[I like Serena]

I did have [1] in mind for [a], but it should work on any generator.

I'm not sure what you are getting at with your last statement.

You mentioned you thought there was a mistake, but you also said I was correct?

I believe the number of homomorphisms is indeed $\gcd(m,n)$, but the formula you gave for ϕ([a]) is not entirely correct.

 

[ArcanaNoir]

So are you saying some of the maps specified by my formula are redundant, some are missing, or some are plain wrong?

 

[I like Serena]

Oh bollocks! I'm saying it should be:
$$\phi([1]) = \frac m {\gcd(m,n)} k[1]$$
with $0 \le k < \gcd(m,n)$.

Can you still smile? :shy:
This should be fun.

 

[ArcanaNoir]

Oh bollocks! I'm saying it should be:
$$\phi([1]) = \frac m {\gcd(m,n)} k[1]$$
with $0 \le k < \gcd(m,n)$.

Can you still smile? :shy:
This should be fun.

Oops. Thanks :)

 

[I like Serena]

Oops. Thanks :)

I guess of the choices offered, it fits "plain wrong".