Centralizer product theorem for elementary abelian group

Statement
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 $$x_i$$.

Generalizations

 * Centralizer product theorem: The same result, except that $$P$$ is now an arbitrary $$p$$-group, rather than an elementary abelian group.

Facts used

 * 1) uses::Maschke's averaging lemma

Proof
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: $$P = C_P(x_1)C_P(x_2) \dots C_P(x_n)$$.

Proof:


 * 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.