Corollary of centralizer product theorem for rank at least three

Statement
Let $$q,r,s$$ be primes such that $$q \ne r, q \ne s$$ (we can have $$r = s$$).

Suppose $$Q$$ is a finite abelian $$q$$-group of rank at least three, $$R$$ is a finite $$r$$-group and $$S$$ is a finite $$s$$-group. Suppose further that $$Q$$ acts as automorphisms on both $$R$$ and $$S$$. Then, there exists a non-identity element $$u \in Q$$ such that $$C_R(u)$$ and $$C_S(u)$$ are both nontrivial.

Similar facts

 * Centralizer product theorem
 * Centralizer product theorem for elementary abelian group

Applications

 * Thompson transitivity theorem

Proof
Given: A prime $$q$$, possibly equal primes $$r,s$$ both distinct from $$q$$. A finite abelian $$q$$-group $$Q$$ of rank at least three (in particular, $$Q$$ is non-cyclic). A finite $$r$$-group $$R$$ and a finite $$s$$-group $$S$$.

To prove: There exists a non-identity element $$u$$ of $$Q$$ such that both $$C_R(u)$$ and $$C_S(u)$$ are nontrivial.

Proof: Without loss of generality, we assume that $$Q$$ is elementary abelian of order $$q^3$$. This is because we can always replace $$Q$$ by an elementary abelian subgroup of order $$q^3$$ in $$Q$$, which exists because $$Q$$ has rank at least three.