Hom, End, and Aut for finite cyclic groups

Statement

Let $m,n\geq1$ be integers and write the cyclic groups additively as $C_m=\mathbb Z/m\mathbb Z$ and $C_n=\mathbb Z/n\mathbb Z$ . Evaluation at $1$ identifies the group homomorphisms with the residues $a\in\mathbb Z/n\mathbb Z$ satisfying $ma=0$ : the corresponding map is $f_a([k]_m)=[ka]_n$ . Thus

|\operatorname{Hom}_{\mathbf{Grp}}(C_m,C_n)|=\gcd(m,n),\qquad \operatorname{End}_{\mathbf{Grp}}(C_n)\cong\mathbb Z/n\mathbb Z, \qquad \operatorname{Aut}_{\mathbf{Grp}}(C_n)\cong(\mathbb Z/n\mathbb Z)^\times.

The middle isomorphism is of rings, with pointwise addition of endomorphisms and composition as multiplication. The last is an isomorphism of groups.

Verification

A homomorphism must send $[k]_m$ to $k f([1]_m)$ . It is well-defined exactly when the relation $m[1]_m=0$ is preserved, which says $n\mid ma$ . Put $d=\gcd(m,n)$ . This divisibility is equivalent to $n/d\mid a$ , giving the $d$ residues $0,n/d,\ldots,(d-1)n/d$ .

For $m=n$ , every residue is allowed and $f_a\circ f_b=f_{ab}$ . Such a map is invertible precisely when $a$ is a unit modulo $n$ . The formula also covers $n=1$ : there is one map and it is the identity on the trivial group.

Forgetting the group structure changes the answer

The underlying sets admit $n^m$ functions from $C_m$ to $C_n$ , while the groups admit only $\gcd(m,n)$ homomorphisms. In particular, $C_2\to C_3$ has nine set maps but just one group map, the zero map.

For one group $C_n$ , the set endomorphisms number $n^n$ and the set automorphisms number $n!$ ; the group endomorphisms number $n$ . The categories of groups, finite groups, and abelian groups give the same homomorphisms on these objects.