Thompson's replacement theorem for abelian subgroups

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.

Similar replacement theorems
For a complete list of replacement theorems, refer:
 * Thompson's replacement theorem for abelian subgroups in arbitrary finite groups
 * Thompson's replacement theorem for elementary abelian subgroups
 * Glauberman's replacement theorem
 * Timmesfeld's replacement theorem

Category:Replacement theorems

Applications

 * Stable version of Thompson's replacement theorem for abelian subgroups: This is a result obtained via repeated application of Thompson's replacement theorem as long as $$B$$ does not normalize the (replaced) $$A$$.
 * Any abelian normal subgroup normalizes an abelian subgroup of maximum order
 * There exists an abelian subgroup of maximum order whose normalizer contains every abelian subgroup it normalizes

Facts used

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

Proof
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 $$N = N_B(A)$$.





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;;

Journal references

 * An extension of Thompson's replacement theorem by algebraic group methods by George Glauberman, Finite Groups 2003
 * An extension of Thompson's replacement theorem by algebraic group methods by George Glauberman, Finite Groups 2003