# 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.

## Facts used

1. 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. Product formula
3. Formula for commutator of element and product of two elements
4. 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 $A \cap M = C_P(A) \cap M = C_M(A)$.
4. $|MC| = |M||C|/|C_M(A)|$: By fact (2), $|MC| = |M||C|/(|C \cap M|)$. By step (3), $C \cap M = C_M(A)$. Thus, we get $|MC| = |M||C|/|C_M(A)|$.
5. 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 $[x,u^{-1}v] \in C_M(A)$.
6. 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 $a \in C_A(M)$.
7. 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.
8. $|A/C| \le |M/C_M(A)|$, hence $|A|/|C| \le |M|/|C_M(A)|$: This follows from the previous step.
9. From steps (4) and (6), we get $|MC| \ge |A|$.