# Commutator subgroup lemma for Frattini-in-center cyclic-center p-group

## Contents

## Statement

Suppose is a non-Abelian -group with the following two properties:

- is a Frattini-in-center group: In other words, is elementary Abelian, or equivalently, , where and are respectively the Frattini subgroup and the center of .
- The center of is a cyclic group.

Then, the following are true:

- is a cyclic subgroup of order . It is the unique minimal normal subgroup as well as the unique minimal characteristic subgroup of .
- Every element of is invariant under all the -automorphisms of . In particular, if is embedded as a normal subgroup of some -group , then .
- If is a normal subgroup of a -group such that , and is a normal subgroup of such that is elementary Abelian, then .

## Related facts

### Applications

## Facts used

- Frattini-in-center p-group implies commutator subgroup is elementary Abelian
- omega-1 of center is normality-large

## Proof

### Proof of part (1)

By fact (1), is elementary Abelian. Also, since is elementary Abelian. Thus, is a nontrivial elementary Abelian subgroup of the cyclic subgroup , forcing to equal , which is cyclic of order .

By fact (2), is normality-large in . Since it is cyclic of prime order, this forces it to be contained in every nontrivial normal subgroup. Thus, it is the unique minimal normal subgroup, and hence also the unique minimal characteristic subgroup.

### Proof of part (2)

Since has order , we cannot have any -automorphisms moving around its non-identity elements. Thus, every element of is invariant under all the -automorphisms of . In particular, if is a normal subgroup of a -group , every element of is invariant under all the inner automorphisms of , so .

(Note: This can also be seen from the fact that is normal in , using that characteristic of normal implies normal, and thus is central, using the fact that minimal normal implies central in nilpotent).

### Proof of part (3)

Since , . In particular, every element of commutes with every element of , and thus, with every element of . This yields that for and , we have:

.

In particular, we have, for and :

.

Since by assumption is elementary Abelian, . We showed in part (2) that , so , so . Thus, , so is generated by elements of order . Since by assumption, we obtain that .