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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:

  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).

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:

  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.

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