# Compute the cardinality of $\mathbf{Hom}(G,C)$

Let $$G$$ be any group and let $$C=\langle a\rangle$$ be a cyclic group (finite or infinite). How many elements does $$\mathbf{Hom}(G,C)$$ have?

The case $$\mathbf{Hom}(C,G)$$ is easy but here I don't know how to proceed. There are $$|\mathbb{Z}|^{|G|}$$ maps, but which of them are true morphisms? If $$G$$ is a finitely generated abelian group, then I can reduce to the cyclic case, but what happens in general?

• You need to first look at $G/[G,G]$, the abelianization of $G$. Any map $G\to C$ will factor through $G/[G,G]$. It will be easier to figure out the morphisms from the abelianization. There may be only one map (the trivial map), for example, if $G=\mathbb{Q}$, or if the order of $G$ is prime to $|a|$. Commented Sep 23, 2019 at 14:50

In general almost anything can happen. Let $$A$$ be an arbitrary set and let $$\mathbb{F}(A)$$ be the free group over $$A$$. Then for any group $$C$$ there's a bijection between $$Hom(\mathbb{F}(A),C)$$ and $$Func(A,C)$$. And so the cardinality $$|C|^{m}$$ is easily achievable for any cardinal number $$m$$. Under the Generalized Continuum Hypothesis this covers most infinite cardinals, except for (so called) limit cardinals. It is an interesting question whether these are achievable (possibly by playing with direct/inverse limits?). I'm not sure. Either way, as you can see $$Hom(G,C)$$ can be arbitrarly large.
On the other extreme we have simple groups. If $$G$$ is a simple group different from $$C$$ then $$Hom(G, C)$$ has exactly one morphism: the zero morphism. Yet another interesting question is whether every finite number is achievable, if we fix $$C$$. I'm not sure about this as well.
Anyway, figuring out morphisms $$G\to C$$ without any knowledge about $$G$$ is definitely doomed to fail.