# Finitely generated implies finitely many homomorphisms to any finite group

This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., finitely generated group) must also satisfy the second group property (i.e., group with finitely many homomorphisms to any finite group)
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## Statement

Suppose $G$ is a finitely generated group. Then, $G$ is a group with finitely many homomorphisms to any finite group. More explicitly, if $K$ is a finite group, then the set of homomorphisms from $G$ to $K$ is finite.

In fact, we have the following explicit upper bound on the number of homomorphisms: if the minimum size of generating set for $G$ is $d$ and the order of $K$ is $n$, then the number of homomorphisms from $G$ to $K$ is $n^d$.

Equality holds in the case that $G$ is a finitely generated free group of rank $d$, though it may also hold in other cases. More generally, equality holds if $G$ admits a quotient that is free of rank $d$ in the subvariety of the variety of groups generated by $K$. For instance, if $K$ is abelian, it suffics for $G$ have a surjective homomorphism to a free abelian group of rank $d$.

## Related facts

### Applications

Statement for group with finitely many homomorphisms to any finite group Corollary for finitely generated group
finitely many homomorphisms to any finite group implies every subgroup of finite index has finitely many automorphic subgroups finitely generated implies every subgroup of finite index has finitely many automorphic subgroups
residually finite and finitely many homomorphisms to any finite group implies Hopfian finitely generated and residually finite implies Hopfian

## Facts used

1. Homomorphism is determined by its restriction to any generating set

## Proof

Given: $G$ is a finitely generated group with minimum size of generating set $d$ (say with $S$ a generating set of minimum size), $K$ is a finite group of order $n$.

To prove: There are at most $n^d$ homomorphisms from $G$ to $K$.

Proof: By Fact (1), to specify a homomorphism from $G$ to $K$, it suffices to specify the restriction to the generating set $S$. Thus, the total number of homomorphisms from $G$ to $K$ is bounded by the number of set maps from $S$ to $K$. By set theory, this number is $n^d$.

### Comment on equality

Note that the bound is achieved if and only if every set map from $S$ to $K$ extends to a homomorphism from $G$ to $K$.