Thompson's lemma on product with centralizer of commutator with abelian subgroup of maximum order

History
This lemma explicitly appears in Gorenstein's text on Finite Groups (see the reference below). It was part of Thompson's proof of his replacement theorem -- however, Thompson did not state it as a separate lemma.

Statement
Suppose $$P$$ is a group of prime power order. Let $$\mathcal{A}(P)$$ denote the set of abelian subgroups of maximum order in $$P$$. Suppose $$A \in \mathcal{A}(P)$$. Suppose $$x \in P$$ is such that the commutator $$[x,A]$$ is an abelian subgroup $$M$$ of $$P$$. Let $$C = C_A(M)$$. Then, $$MC \in \mathcal{A}(P)$$.

The operation sending $$A$$ to the new subgroup $$MC$$ is termed (on this wiki) the Thompson replacement operation.

Applications

 * Thompson's replacement theorem for abelian subgroups
 * Thompson's replacement theorem for elementary abelian subgroups

Facts used

 * 1) uses::Equivalence of definitions of maximal among abelian subgroups: An abelian subgroup of a group that is not contained in any bigger abelian subgroup is a self-centralizing subgroup: it equals its own centralizer.
 * 2) uses::Product formula
 * 3) uses::Formula for commutator of element and product of two elements
 * 4) uses::Witt's identity

Proof
Given: A group $$P$$ of prime power order. $$\mathcal{A}(P)$$ is the set of abelian subgroups of maximum order in $$P$$. A subgroup $$A \in \mathcal{A}(P)$$. An element in $$x \in P$$ such that $$M = [x,A]$$ is abelian. $$C = C_A(M)$$.

To prove: $$MC \in \mathcal{A}(P)$$.

Proof:


 * 1) $$MC$$ is abelian: By assumption, $$M$$ is abelian. Since $$C = C_A(M) \le A$$, $$C$$ is also abelian. Further, every element of $$M$$ commutes with every element of $$C$$ by definition, so $$MC$$ is abelian.
 * 2) $$C_P(A) = A$$: Since $$A$$ is abelian of maximum order, it is maximal among abelian subgroups: it is not contained in a bigger abelian subgroup. Thus, by fact (1), $$C_P(A) = A$$.
 * 3) $$C \cap M = A \cap M = C_M(A)$$: Clearly, since $$C \le A$$, $$C \cap M \le A \cap M$$. Conversely, anything in $$A \cap M$$ is in particular inside $$M$$, so centralizes $$M$$, hence is in $$A \cap C_P(M) = C$$. Thus, $$A \cap M \le C \cap M$$, and we get $$C \cap M = A \cap M$$. Finally, since step (2) yields $$C_P(A) = A$$, we get $$A \cap M = C_P(A) \cap M = C_M(A)$$.
 * $$|MC| = |M||C|/|C_M(A)|$$: By fact (2), $$|MC| = |M||C|/(|C \cap M|)$$. By step (3), $$C \cap M = C_M(A)$$. Thus, we get $$|MC| = |M||C|/|C_M(A)|$$.
 * 1) Consider the map $$a \mapsto [x,a]$$ from $$A$$ to $$M = [x,A]$$. Then, if $$[x,u]$$ and $$[x,v]$$ are in the same coset of $$C_M(A)$$, then $$[x,u^{-1}v] \in C_M(A)$$: Suppose $$[x,u]$$ and $$[x,v]$$ are in the same coset of $$C_M(A)$$ in $$M$$. Then, by fact (3), $$[x,v][x,u]^{-1} = c_u([x,u^{-1}v])$$. Since the left side is in $$C_M(A)$$, so is the right side. By definition of $$C_M(A)$$, we get that $$[x,u^{-1}v] \in C_M(A)$$.
 * 2) If $$a \in A$$ and $$[x,a] \in C_M(A)$$, then $$a \in C_A(M)$$: If $$[x,a] \in C_M(A)$$, then $$[[x,a],b]$$ is the identity for all $$b \in A$$. By fact (4), and the abelianness of $$A$$, we obtain that $$[[x,c],a]$$ is the identity for all $$c \in A$$. In particular, $$a$$ commutes with $$M$$, so $$a \in C_A(M)$$.
 * 3) The map $$a \mapsto [x,a]$$ from $$A$$ to $$M$$ has the property that if $$[x,u]$$ and $$[x,v]$$ are in the same coset of $$C_M(A)$$, then $$u,v$$ are in the same coset of $$C_A(M)$$. Thus, it is an injective map from $$A/C_A(M)$$ to $$M/C_M(A)$$: This follows from the previous two steps.
 * $$|A/C| \le |M/C_M(A)|$$, hence $$|A|/|C| \le |M|/|C_M(A)|$$: This follows from the previous step.
 * 1) From steps (4) and (6), we get $$|MC| \ge |A|$$.