Thompson's replacement theorem for abelian subgroups

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This article defines a replacement theorem
View a complete list of replacement theorems| View a complete list of failures of replacement

Statement

Suppose P is a group of prime power order (we'll call the prime p).

Let \mathcal{A}(P) denote the set of all abelian subgroups of maximum order in P (i.e., |A| \ge |B| for all abelian subgroups B of P).

Suppose A \in \mathcal{A}(P) and B is an abelian subgroup such that A normalizes B but B does not normalize A. Then, there exists an abelian subgroup A^* of AB such that:

  1. A^* \in \mathcal{A}(P).
  2. A \cap B is a proper subgroup of A^* \cap B.
  3. A^* is contained in \langle A^B \rangle = \langle A^{AB} \rangle, i.e., it is contained in the normal closure of A in AB. In particular, it is contained in the normal closure \langle A^P \rangle of A in P.
  4. A^* normalizes A.

We have the following additional conclusions:

  1. A^* normalizes B: Note that since A^* is contained in AB, and AB \le N_P(B) by assumption, we see that A^* normalizes B as well.
  2. A^* \cap B is a proper subgroup of B: This is because A^* cannot contain B, since A^* normalizes A but B does not normalize A.

Thus, all the conditions assumed for A also hold for A^*, except possibly the fact that B does not normalize A. Hence, the term replacement.

Related facts

Similar replacement theorems

For a complete list of replacement theorems, refer:

Category:Replacement theorems

Applications

Facts used

  1. Thompson's lemma on product with centralizer of commutator with abelian subgroup of maximum order
  2. Nilpotent implies center is normality-large
  3. Normality satisfies image condition

Proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

Given: A finite p-group P. \mathcal{A}(P) is the set of abelian subgroups of P of maximum order. A \in \mathcal{A}(P), and B is a subgroup of A such that A normalizes B but B does not normalize A.

To prove: There exists A^* \in \mathcal{A}(P) such that A \cap B is a proper subgroup of A^* \cap B, A^* normalizes A and B, and A^* is contained in AB.

Proof: We let C = AB and N = N_B(A).

basic diagram of key initial constructions (steps 1-6)
More detailed diagram
Step no. Assertion Facts used Given data used Proof steps used Explanation
1 C = AB is a subgroup of P, and B is normal in C -- A normalizes B -- [SHOW MORE]
2 B normalizes N -- B is abelian -- [SHOW MORE]
3 A normalizes N -- A normalizes B -- [SHOW MORE]
4 N = N_B(A) is normal in C -- -- steps (2), (3) [SHOW MORE]
5 N is a proper subgroup of B -- B does not normalize A [SHOW MORE]
6 B/N is a nontrivial normal subgroup of C/N fact (3) -- steps (1), (4), (5) [SHOW MORE]
7 B/N \cap Z(C/N) is nontrivial fact (2) -- step (6) Direct combination of the facts
8 Let x \in P be such that its image mod N is a nontrivial element in B/N \cap Z(C/N). Then M = [x,A] is contained in N and hence in B step (7) Note that step (7) guarantees the existence of such x. [SHOW MORE]
9 M = [x,A] is an abelian subgroup of P B is abelian step (8) [SHOW MORE]
10 A^* = MC_A(M) is an abelian subgroup of maximum order in P fact (1) A is abelian of maximum order step (9) Piece together.
11 A \cap B is contained in C_A(M) B is abelian step (8) [SHOW MORE]
12 A \cap B is contained in A^* \cap B -- step (11) [SHOW MORE]
13 A^* normalizes A step (8) [SHOW MORE]
14 M is not contained in A step (8) [SHOW MORE]
15 A^* \cap B is not contained in A \cap B steps (8), (14) [SHOW MORE]
16 A^* is contained in C = AB steps (1), (8), (10) [SHOW MORE]
17 A^* is contained in \langle A^C \rangle. steps (8), (9), (10) [SHOW MORE]
18 A^* normalizes B step (1), (16) [SHOW MORE]

The conclusions of steps (10), (12), (13), (15), (16), (17), and (18) complete the proof.

Conceptual summary of proof

The key idea of the proof is to use fact (1) somehow, and this requires finding a suitable x such that [x,A] is abelian. and the resulting subgroup A^* that we get is nice enough.

Working backward from this, we see that we would like that [x,A] not be a subgroup of A, which in turn means that we want x outside N_P(A). On the other hand, we do want A^* to normalize A, which means that we want [x,A] to lie inside N_P(A). Finally, we would like [x,A] \subseteq B so that the subsequent group constructed is in AB. With all these considerations, we should aim to find x outside N_B(A) but inside B, but such that [x,A] \subseteq N_B(A).

This motivation now allows us to work out the proof details given above. Here is a rough outline of the proof:

  • Steps (1)-(6) set things up. The fact that B does not normalize A is used to show that B/N is a nontrivial normal subgroup of P/N.
  • The crucial steps for the construction are (7) and (8), that involve the choice of M. Steps (9) and (10) now use fact (1) to get a hold on A^*.
  • The remaining steps help us get a clearer idea of the subgroup inclusions and prove the remaining desired properties for A^*.

GAP implementation

Here is a GAP implementation of the constructive approach used in the proof:

ThompsonsReplacementAbelian := function(P,A,B)
        local C,N,x,M,CAM,Astar;
        C := Group(Union(A,B));
        N := Normalizer(B,A);
        x := Filtered(Set(B),y -> (not(y in N)) and IsSubgroup(N,CommutatorSubgroup(Group(y),C)))[1];
        M := Group(List(Set(A),y -> x^(-1) * y^(-1) * x * y));
        CAM := Centralizer(A,M);
        Astar := Group(Union(M,CAM));
        return Astar;
end;;

References

Textbook references

Journal references