Centralizer product theorem for elementary abelian group

From Groupprops
Jump to: navigation, search


Suppose p,q are distinct primes. Suppose P is an elementary abelian p-group, and Q \le \operatorname{Aut}(P) is an abelian q-group that is not cyclic. Let x_1, x_2, \dots, x_n be an enumeration of the non-identity elements of Q. Then, we have:

P = C_P(x_1)C_P(x_2) \dots C_P(x_n).

where C_P(x_i) denotes the set of fixed points in P of the automorphism x_i.

Related facts


Facts used

  1. Maschke's averaging lemma


Given: Primes p \ne q. An elementary abelian p-group P, a non-cyclic abelian q-subgroup Q of \operatorname{Aut}(P). x_1, x_2, \dots, x_n are the non-identity elements of Q.

To prove: P = C_P(x_1)C_P(x_2) \dots C_P(x_n).


  1. By fact (1), viewing P as a vector space over the field of p elements and Q as a group acting on this vector space, we can decompose P as a direct sum of irreducible Q-modules, say P_1, P_2, \dots, P_r.
  2. Let Q_i be the pointwise stabilizer of P_i in Q. Then, Q/Q_i is cyclic: Q/Q_i is an abelian group of automorphisms of the vector space P_i, and P_i is irreducible. This forces Q/Q_i to be cyclic.
  3. Each Q_i is a nontrivial subgroup of Q: This follows from the previous step and the fact that Q isn't cyclic.
  4. For each i, pick a non-identity element x of Q_i. Then, C_P(x) contains P_i: This follows from the definition of Q_i.
  5. P = \sum C_P(x_i): Each P_i is contained in C_P(x) for some non-identity element x. Thus, P is the sum of all C_P(x), x ranging over the non-identity elements of P.

Reverting to multiplicative notation yields the result.


Textbook references

  • Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page 69, Theorem 3.3, Chapter 3, Section 3 (Complete reducibility), More info