# Normal and self-centralizing implies coprime 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., self-centralizing normal subgroup) must also satisfy the second subgroup property (i.e., coprime automorphism-faithful normal subgroup)

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This article states and (possibly) proves a fact about a finite group and a Coprime automorphism group (?): a subgroup of the automorphism group whose order is relatively prime to the order of the group itself.

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## Contents

## Statement

### Verbal statement

Any normal Self-centralizing subgroup (?) of a finite group is coprime automorphism-faithful.

### Statement with symbols

Suppose is a finite group and is a normal subgroup of such that (or equivalently, ). Then, if is a non-identity automorphism of whose order is relatively prime to the order of , and if , then the restriction of to is a non-identity automorphism of .

## Related facts

### Similar facts

- Characteristic and self-centralizing implies coprime automorphism-faithful: Note that when the subgroup is characteristic, then any automorphism
*does*restrict to an automorphism of the subgroup, so the statement can be reformulated somewhat. - Normal and centralizer-free implies automorphism-faithful: The proof idea is the same, but the stronger assumption yields a stronger conclusion.

### Opposite facts

- Normal and self-centralizing implies coprime automorphism-faithful in infinite p-group: The analogous result also holds for an infinite p-group.

## Facts used

## Proof

### Hands-on proof

**Given**: A finite group , a normal subgroup such that . An automorphism of that acts as identity on , and such that the order of is relatively prime to the order of

**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 by assumption, we obtain that , and thus, and are in the same coset of in . Thus, stabilizes the subnormal series:

.

Thus, by fact (2), must be the identity automorphism.

### Proof using three subgroup lemma

Note that this proof superficially appears different from the preceding proof, but the part of the proof using the three subgroup lemma is precisely equivalent to the initial part of the proof above upto the point where we deduce that . Thinking of the proof in terms of the three subgroup lemma, rather than in terms of step-by-step manipulations as outlined above, is more useful in some situations.

**Given**: A finite group , a normal subgroup such that . An automorphism of that acts as identity on , and such that the order of is relatively prime to the order of

**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 the three subgroup lemma, is trivial. Since , we see that , so by the assumption of self-centralizing, we obtain that . Thus, stabilizes the subnormal series:

Now, using the fact that the stability group of the subnormal series of a finite group has no other prime factors, we conclude that must be the trivial group, so is the identity automorphism.