Central product decomposition lemma for characteristic rank one

Statement
Suppose $$G$$ is a finite p-group of characteristic rank one, i.e., $$G$$ is a group of prime power order and every Abelian characteristic subgroup of $$G$$ is cyclic. Then, if $$C$$ is a critical subgroup of $$G$$, $$Z(C)$$ is a cyclic group. Further, there exists a subgroup $$E$$ of $$C$$ satisfying the following conditions:


 * 1) $$E$$ is an extraspecial group
 * 2) $$E$$ is a central factor of $$G$$
 * 3) $$Z(C)E = C$$ (i.e., $$E$$ is a cocentral subgroup of $$G$$).
 * 4) If $$R = C_G(E)$$ denotes the centralizer of $$E$$ in $$G$$, then $$Z(C)$$ is a self-centralizing subgroup of $$R$$

(Note that the existence of critical subgroups is guaranteed by Thompson's critical subgroup theorem).

Facts used

 * 1) uses::Characteristic rank one is characteristic subgroup-closed
 * 2) uses::Characteristic rank one implies cyclic-center
 * 3) uses::Extraspecial commutator-in-center subgroup is central factor

Critical subgroup has a cyclic center
Since $$G$$ has characteristic rank one, and $$C$$ is a characteristic subgroup of $$G$$, $$C$$ also has characteristic rank one (fact (1)). Thus, the center of $$C$$ is cyclic, so $$Z(C)$$ is a cyclic group.

Construction of $$E$$
We consider three cases:


 * 1) $$C$$ is Abelian, so $$Z(C) = C$$: In this case, we can take $$E$$ to be trivial
 * 2) $$C$$ is extraspecial: In this case, we can take $$E = C$$
 * 3) $$C$$ is neither Abelian nor extraspecial

Let's study case (3) in more detail. Since $$C$$ has class two, we see that for any $$x,y \in C$$, $$[x,y^p] = [x,y]^p$$. Further, since $$C/Z(C)$$ is elementary Abelian, $$y^p \in Z(C)$$, so $$[x,y^p]$$ is the identity. Thus, $$[x,y]$$ has order $$p$$, so $$C' \le \Omega_1(Z(C))$$.

On the other hand, since $$Z(C)$$ is cyclic, $$\Omega_1(Z(C))$$ is of order $$p$$. And since $$C$$ is non-Abelian, $$C'$$ is nontrivial. Thus, $$C' = \Omega_1(Z(C))$$ is of order $$p$$.

Now, we go modulo $$C'$$. Note that since $$C$$ is not extraspecial, $$C'$$ is not equal to $$Z(C)$$. Let $$\overline{Z} = Z(C)/C'$$, and $$\overline{C} = C/C'$$. Then, $$\overline{C}$$ is an Abelian group, with the property that all the elements of order $$p$$ are in $$\overline{Z}$$. By the structure theory of Abelian $$p$$-groups, we can write:

$$\overline{C} = \overline{A} \times \overline{E}$$

where $$\overline{A}$$ is a cyclic subgroup containing $$\overline{Z}$$, and $$\overline{E}$$ is an elementary Abelian $$p$$-group. Further, the index of $$\overline{Z}$$ in $$\overline{A}$$ is either 1 or $$p$$.

(note that the choice of $$\overline{A}$$ and $$\overline{E}$$ is not unique).

Let $$A,E$$ be the inverse images mod $$C'$$ of $$\overline{A}$$ and $$\overline{E}$$. Our goal is to show that the $$E$$ that we've constructed in this manner satisfies the four specified conditions.

Proof that it is extraspecial
Let $$D = Z(C)E$$. We first observe that $$D$$ is characteristic in $$C$$. Indeed, if $$\overline{A} = \overline{Z}$$, then $$D = C$$, so it is characteristic. Otherwise, $$D$$ is the product of $$\Omega_1(C)$$ and $$\operatorname{Agemo}^1(C)$$, which is again characteristic in $$C$$. In either case, since $$C$$ has characteristic rank one, so does $$D$$, so the center $$Z(D)$$ is cyclic. But since $$Z(C) \le D \le C$$, we have $$Z(C) \le Z(D)$$, so the center of $$D$$ is a cyclic subgroup containing $$Z(C)$$. But by our construction, $$Z(C)$$ is maximal among cyclic subgroups of $$D$$, so $$Z(D) = Z(C)$$.

But we also have, since $$D$$ is a product of $$E$$ and $$Z(C)$$ (which commutes with $$E$$), that $$Z(E) \le Z(D)$$. Hence $$Z(E) \le Z(C) \cap E = C'$$, and since $$Z(E)$$ is nontrivial, $$Z(E) = C'$$. Further, it's clear that $$E$$ properly contains $$Z(E)$$, otherwise $$C$$ would be cyclic. Thus, $$E'$$ is nontrivial. But $$E' \le C'$$, and $$E$$ is not Abelian, so $$E' = C' = Z(E)$$. Finally $$E/E' = E/C' = \overline{E}$$ is elementary Abelian by our construction. Thus, $$E$$ is extraspecial with center equal to $$C'$$.

Proof that it is a central factor of the whole group
By the definition of critical subgroup, $$[G,C] \le Z(C)$$. In particular, $$[G,E] \le Z(C)$$. We want to show that $$[G,E] \le Z(E)$$. Recall for this that $$Z(E) = E' = C' = \Omega_1(Z(C))$$, so what we need to show is that every element of $$[G,E]$$ has order $$p$$.

Let's see that. If $$g \in G$$ and $$h,k \in E$$, then $$[g,k]$$, being in $$Z(C)$$, commutes with $$h \in E \le C$$. Thus, we have that the commutator with $$g$$ is an endomorphism:

$$[g,h][g,k] = [g,hk]$$.

In particular, $$[g,h^p]= [g,h]^p$$ for $$g \in G, h \in E$$. By construction, $$E$$ being extraspecial, $$h^p \in Z(E)$$. Now, $$Z(G)$$ is a nontrivial subgroup of $$C$$ contained in $$Z(C)$$, and hence must contain $$Z(E)$$. In particular, $$h^p \in Z(G)$$, so $$[g,h^p]$$ is trivial, so $$[g,h]$$ has order dividing $$p$$. In particular, all commutators are in $$Z(E)$$, so $$[G,E] \le Z(E)$$.

The proof now follows from fact (3).

Proof that it is cocentral in the critical subgroup
Let $$R = C_G(E)$$. We have established that $$G = ER$$. In particular, $$C = E(R \cap C)$$. Also, clearly $$Z(C) \le R$$, and $$R \cap E = Z(E) = \Omega_1(Z(C)) = E' = C'$$. We want to show that $$Z(C) = R \cap C$$.

Textbook references

 * , Page 196, Lemma 4.7 (Section 5.4)