# Finitely many homomorphisms to any finite group implies every subgroup of finite index has finitely many automorphic subgroups

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., group with finitely many homomorphisms to any finite group) must also satisfy the second group property (i.e., group in which every subgroup of finite index has finitely many automorphic subgroups)
View all group property implications | View all group property non-implications
Get more facts about group with finitely many homomorphisms to any finite group|Get more facts about group in which every subgroup of finite index has finitely many automorphic subgroups

## Statement

Suppose $G$ is a group with finitely many homomorphisms to any finite group. Then, $G$ is also a group in which every subgroup of finite index has finitely many automorphic subgroups.

## Definitions used

For convenience, we use the following definitions. For more on why these definitions are equivalent to other definitions, see equivalence of definitions of group with finitely many homomorphisms to any finite group and equivalence of definitions of group in which every subgroup of finite index has finitely many automorphic subgroups:

Term Definition
group with finitely many homomorphisms to any finite group For any natural number $n$, there are only finitely many normal subgroups of the group of index $n$
group in which every subgroup of finite index has finitely many automorphic subgroups Any normal subgroup of finite index has only finitely many automorphic subgroups.

## Proof

The proof is immediate with the chosen definitions.

Given: A group $G$ with the property that for any natural number, there are only finitely many normal subgroups of $G$ of index equal to that natural number. A normal subgroup $H$ of $G$ of index $n$.

To prove: There are only finitely many subgroups $K$ of $G$ for which there exists an automorphism $\sigma$ of $G$ satisfying $\sigma(H) = K$.

Proof:

Step no. Assertion/construction Given data used Previous steps used Explanation
1 For any automorphism $\sigma$ of $G$, $\sigma(H)$ is a normal subgroup of index $n$ in $G$. $H$ is normal of index $n$ in $G$. -- automorphisms preserve normality and index of subgroups
2 There are only finitely many subgroups $K$ of $G$ for which there exists an automorphism $\sigma$ of $G$ satisfying $\sigma(H) = K$. $G$ has only finitely many normal subgroups of index $n$. Step (1) Step-combination direct