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

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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.

Related facts

Applications

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|.

References

Textbook references

Journal references