# Centerless and characteristic in automorphism group implies automorphism group is complete

## Contents

## Statement

Suppose is a Centerless group (?) such that, with the natural embedding of in its Automorphism group (?) , is a Characteristic subgroup (?). Then, the automorphism group is a Complete group (?): it is centerless and every automorphism of it is inner.

## Related facts

- Centerless implies inner automorphism group is centralizer-free in automorphism group: This states that if is centerless, is trivial. In particular, it shows that is centerless. This idea leads to the notion of an automorphism tower.
- Wielandt's automorphism tower theorem: This states that for any finite centerless group, the tower obtained by repeatedly taking automorphism groups eventually stabilizes. Note that our result gives a condition under which it stabilizes after just one step: the automorphism group is itself complete, so the tower stabilizes after that.

## Facts used

### For the hands-on proof

- Centerless implies inner automorphism group is centralizer-free in automorphism group: If is centerless, then (identified naturally with ) has trivial centralizer in . In other words, no non-identity automorphism commutes with every inner automorphism.
- Inner automorphism group is normal in automorphism group
- Centralizer-free and normal implies automorphism-faithful: If is a normal subgroup and is trivial, then any automorphism of that fixes every element of is the identity map.
- The basic idea that, if denotes conjugation by and denotes an automorphism of , we have:

In other words, if is an automorphism of , the action of on by conjugation in precisely mimics the action of as an automorphism on .

### For the proof using the automorphism group action lemma

Automorphism group action lemma: Suppose is a group and are subgroups such that and is an automorphism of that restricts to an automorphism of as well as of . Let be the induced automorphism of and be the induced automorphism of . Then, if

denotes the homomorphism induced by the action by conjugation, we have:

.

Here, denotes conjugation by in the group .

## Proof

### Hands-on proof

**Given**: A centerless group , embedded naturally as in the automorphism group . Further, is a characteristic subgroup of under this embedding.

**To prove**: is complete: the center of is trivial, and any automorphism of is inner.

**Proof**:

- (
**Facts used**: fact (1)): is centerless: By fact (1), is centralizer-free in . Thus, is centerless. - (
**Facts used**: facts (1), (2), (3)): is automorphism-faithful in : By fact (1), is centralizer-free in , and by fact (2), is normal in . Combining these with fact (3), we obtain that is automorphism-faithful in . - (
**Given data used**: is characteristic in ): For any automorphism of , there exists an element such that conjugation by has the same effect as on the subgroup : First, observe that since is characteristic in , the restriction of to equals an automorphism of , so there is an automorphism of having the same effect. Note that (by fact (4)) an automorphism of manifests itself as the inner automorphism by on . Thus, and have the same effect on , viewed as a subgroup of . - , and thus, is inner: By step (3), is an automorphism of whose restriction to is the identity map. Step (2) now forces to be the identity map on the whole group , yielding .
- Step (1) shows that is centerless, and steps (3) and (4) show that every automorphism is inner. This completes the proof of completeness.

### Proof using automorphism group action lemma

**Given**: A centerless group , embedded naturally as in the automorphism group . Further, is a characteristic subgroup of under this embedding.

**To prove**: is complete: the center of is trivial, and any automorphism of is inner.

**Proof**: Set and . By assumption, is characteristic in , so . Let be any automorphism of . Observe that:

- restricts to an automorphism of , because .
- restricts to an automorphism of , because by assumption is characteristic in .

Further, the natural map

is in this case the identity map. Thus, the automorphism group action lemma tells us that for any automorphism of , we have:

where is the automorphism of induced by .