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?

  • 3
    $\begingroup$ 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 $|a|$. $\endgroup$ Commented Sep 23, 2019 at 14:50

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .