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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Contents
Statement
Suppose is a finite p-group of characteristic rank one, i.e., is a group of prime power order and every Abelian characteristic subgroup of is cyclic. Then, if is a critical subgroup of , is a cyclic group. Further, there exists a subgroup of satisfying the following conditions:
- is an extraspecial group
- is a central factor of
- (i.e., is a cocentral subgroup of ).
- If denotes the centralizer of in , then is a self-centralizing subgroup of
(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 has characteristic rank one, and is a characteristic subgroup of , also has characteristic rank one (fact (1)). Thus, the center of is cyclic, so is a cyclic group.
Construction of
We consider three cases:
- is Abelian, so : In this case, we can take to be trivial
- is extraspecial: In this case, we can take
- is neither Abelian nor extraspecial
Let's study case (3) in more detail. Since has class two, we see that for any , . Further, since is elementary Abelian, , so is the identity. Thus, has order , so .
On the other hand, since is cyclic, is of order . And since is non-Abelian, is nontrivial. Thus, is of order .
Now, we go modulo . Note that since is not extraspecial, is not equal to . Let , and . Then, is an Abelian group, with the property that all the elements of order are in . By the structure theory of Abelian -groups, we can write:
where is a cyclic subgroup containing , and is an elementary Abelian -group. Further, the index of in is either 1 or .
(note that the choice of and is not unique).
Let be the inverse images mod of and . Our goal is to show that the that we've constructed in this manner satisfies the four specified conditions.
Proof that it is extraspecial
Let . We first observe that is characteristic in . Indeed, if , then , so it is characteristic. Otherwise, is the product of and , which is again characteristic in . In either case, since has characteristic rank one, so does , so the center is cyclic. But since , we have , so the center of is a cyclic subgroup containing . But by our construction, is maximal among cyclic subgroups of , so .
But we also have, since is a product of and (which commutes with ), that . Hence , and since is nontrivial, . Further, it's clear that properly contains , otherwise would be cyclic. Thus, is nontrivial. But , and is not Abelian, so . Finally is elementary Abelian by our construction. Thus, is extraspecial with center equal to .
Proof that it is a central factor of the whole group
By the definition of critical subgroup, . In particular, . We want to show that . Recall for this that , so what we need to show is that every element of has order .
Let's see that. If and , then , being in , commutes with . Thus, we have that the commutator with is an endomorphism:
.
In particular, for . By construction, being extraspecial, . Now, is a nontrivial subgroup of contained in , and hence must contain . In particular, , so is trivial, so has order dividing . In particular, all commutators are in , so .
The proof now follows from fact (3).
Proof that it is cocentral in the critical subgroup
Let . We have established that . In particular, . Also, clearly , and . We want to show that .
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Textbook references
- Finite Groups by Daniel Gorenstein, ISBN 0821843427, ^{More info}, Page 196, Lemma 4.7 (Section 5.4)