Central product decomposition lemma for characteristic rank one

This article gives the statement, and possibly proof, of a particular subgroup of kind of subgroup in a group being self-centralizing. In other words, the centralizer of the subgroup in the group is contained in the subgroup
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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:

$E$ is an extraspecial group
$E$ is a central factor of $G$
$Z(C)E=C$ (i.e., $E$ is a cocentral subgroup of $G$ ).
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).

Definitions used

Critical subgroup

Central factor

Facts used

Proof

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:

$C$ is Abelian, so $Z(C)=C$ : In this case, we can take $E$ to be trivial
$C$ is extraspecial: In this case, we can take $E=C$
$C$ 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 $C'\leq \Omega _{1}(Z(C))$ .

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 $Z(C)\leq D\leq C$ , we have $Z(C)\leq Z(D)$ , 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 $Z(E)\leq Z(D)$ . Hence $Z(E)\leq Z(C)\cap E=C'$ , 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 $E'\leq C'$ , 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, $[G,C]\leq Z(C)$ . In particular, $[G,E]\leq Z(C)$ . We want to show that $[G,E]\leq Z(E)$ . 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 $h\in E\leq C$ . 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 $[G,E]\leq Z(E)$ .

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 $Z(C)\leq R$ , and $R\cap E=Z(E)=\Omega _{1}(Z(C))=E'=C'$ . We want to show that $Z(C)=R\cap C$ .