Let $G$ be a finite abelian group of order n. How many distinct group homomorphisms are there from $G$ to $R/Z$?

Question

First, is this question well-defined? That is, the homomorphisms are determined by the order n.

Since $R/Z$ is the multiplicative unit circle, all finite subgroups are cyclic, so the image of the homomorphisms must be cyclic. My intuition is there exists bijection between cyclic subgroups of $G$, but I don't know how to proceed it formally.

Answer 1

This gets you into group cohomology.

For any subgroup $A\le G$ such that $G/A$ is cyclic, it's the number of central extensions $$0\to A\to G\to G/A\to 0,$$ by the first isomorphism theorem.

These are at least as many as in the second cohomology group, $H^2(G/A,A).$ (When the action of $G/A$ on $A$ is trivial, the correspondence is one-one.)

In the situation where the extension is split we have $G\cong A×G/A,$ and this corresponds to zero in $H^2(G/A,A).$

Answer 2

The question does indeed turn out to be well-defined, but this is not obvious and the "order $n$" part makes it a little confusing anyway as it's much easier to answer for all $n$ at once.

If $G$ is cyclic of order $m$, let $t$ be a generator. Then $t$ has to go to something of the form $i/m$ (mod $\mathbb{Z})$, since these are the only order $m$ elements of the circle, and where $t$ goes determines the whole map, so there are $m$ maps.

Now we can bootstrap from the cyclic case as follows. If $G = G_1 \oplus G_2$, then $\text{Hom}(G, A) = \text{Hom}(G_1, A) \times \text{Hom}(G_2, A)$ for any abelian group $A$.

So if we've proven the result for cyclic groups that $| G | = | \text{Hom}(G, \mathbb{R}/\mathbb{Z})|$, we deduce it for abelian groups that are products of two distinct cyclic groups (then inductively three, four, ...). Since the fundamental theorem of finitely generated abelian groups tells you your group is a product of some number of cyclic groups, you are done.