# Central product decomposition lemma for characteristic rank one

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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:

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).

## 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:

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 $Z(C) = R \cap C$.