# Characteristic implies automorph-conjugate

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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., characteristic subgroup) must also satisfy the second subgroup property (i.e., automorph-conjugate subgroup)
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## Statement

### Symbolic statement

Let $H$ be a characteristic subgroup of $G$. Then $H$ is an automorph-conjugate subgroup of $G$.

### Property-theoretic statement

The subgroup property of being characteristic is stronger than the subgroup property of being automorph-connjugate.

## Definitions used

### Characteristic subgroup

Further information: Characteristic subgroup

A subgroup of a group is termed a characteristic subgroup if any automorphism of the group maps the subgroup to itself.

That is, $H \le G$ is characteristic if for any automorphism $\sigma$ of $G$, $\sigma(H)=H$.

Characteristicity can also be expressed using the relation implication expression:

Automorphic subgroups $\implies$ Equal subgroups

In other words, any subgroup obtained as $\sigma(H)$ where $\sigma$ is an automorphism of $G$, is equal to $H$.

### Automorph-conjugate subgroup

Further information: Automorph-conjugate subgroup A subgroup of a group is termed an automorph-conjugate subgroup if any automorphism of the group maps the subgroup to a conjugate subgroup.

That is $H \le G$ is automorph-conjugate if for any automorphism $\sigma$ of $G$ there exists $g \in G$ such that $\sigma(H) = gHg^{-1}$ (the latter is sometimes denoted $H^g$).

This property can also be expressed using the relation implication expression:

Automorphic subgroups $\implies$ Conjugate subgroups

In other words, any subgroup obtained as $\sigma(H)$ where $\sigma$ is an automorphism of $G$, is also a conjugate subgroup to $H$.

## Proof

### Hands-on proof

Given: $H$ is a characteristic subgroup of $G$,

To prove: $H$ is an automorph-conjugate subgroup of $G$. In other words, for any automorphism $\sigma$ of $G$, there exists $g \in G$, $\sigma(H) = H^g$.

Proof: For any automorphism $\sigma$ of $G$, $\sigma(H) = H$ (by definition of characteristicity). Clearly $H$ is a conjugate subgroup to itself (say $H^e = H$ where $e$ is the identity element). Thus, $H$ is automorph-conjugate.

### Using relation implication expressions

This subgroup property implication can be proved using relation implication expressions for the subgroup properties
View other implications proved in this way OR Read a survey article on the topic

The property of being a characteristic subgroup can be expressed as a relation implication:

Automorphic subgroups $\implies$ Equal subgroups

The property of being an automorph-conjugate subgroup can be expressed as a relation implication:

Automorphic subgroups $\implies$ Conjugate subgroups

Since the left side is the same for both properties, but the right side is stronger for characteristicity, we see that characteristic implies automorph-conjugate.

## Converse

The converse is not true. Counterexamples can be easily constructed by looking at Sylow subgroups which are not normal. This is because any Sylow subgroup must be an isomorph-conjugate subgroup, and hence an automorph-conjugate subgroup.