# Normal and centralizer-free implies automorphism-faithful

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., centralizer-free normal subgroup) must also satisfy the second subgroup property (i.e., automorphism-faithful normal subgroup)

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## Contents

## Statement

Suppose is a subgroup of a group such that:

- is centralizer-free in : The centralizer is trivial.
- is normal in .

Then, is an Automorphism-faithful subgroup (?) of : every nontrivial automorphism of that restricts to must restrict to a nontrivial automorphism of .

## Related facts

### Analogues in other algebraic structures

- Centralizer-free ideal implies automorphism-faithful
- Centralizer-free ideal implies derivation-faithful

## Facts used

## Proof

### Hands-on proof

**Given**: A group , a normal subgroup such that is trivial. An automorphism of that acts as the identity on .

**To prove**: is the identity automorphism.

**Proof**: Note that for , we have, by normality, that . Thus, since acts as the identity on , we get:

.

Expand the left side and use that to obtain:

.

Rearranging this yields that commutes with . Since this holds true for every , we obtain that . Since is trivial, we obtain that for every , is the identity element, yielding for all . This yields that is the identity automorphism.

### Proof using three subgroup lemma

Note that although this proof may superficially appear different from the preceding one, it is a less hands-on formulation of the same proof. This style of proof may be more useful when proving similar statements that are more complicated; for instance: normal and self-centralizing implies coprime automorphism-faithful.

**Given**: A group , a normal subgroup such that is trivial. An automorphism of that acts as identity on .

**To prove**: is the identity automorphism.

**Proof**: Suppose is the subgroup generated by in , and let's assume we're working in . By assumption, is trivial. Since is normal in , we have . Thus, we have:

- , which is trivial, so is trivial
- is trivial, because is trivial

So, by fact (1) (the three subgroup lemma), is trivial. Since , we see that , so by the assumption of being trivial, we obtain that is trivial. This translates to saying that every element of fixes every element of , so must be the identity automorphism.