Congruence condition on number of Sylow subgroups containing a given subgroup of prime power order

Statement
Suppose $$G$$ is a finite group and $$p$$ is a prime number. Suppose $$A$$ is a $$p$$-subgroup of $$G$$, i.e., a subgroup of $$G$$ whose order is a power of $$p$$. Then, the number of $$p$$-fact about::Sylow subgroups of $$G$$ containing $$A$$ is congruent to $$1$$ modulo $$p$$.

Related facts

 * Congruence condition on Sylow numbers
 * Congruence condition on number of subgroups of given prime power order

Facts used

 * 1) uses::Fundamental theorem of group actions. See also uses::class equation of a group action, which can be used directly in place of the fundamental theorem to shorten the proof.
 * 2) uses::Congruence condition on Sylow numbers: This states that the total number of $$p$$-Sylow subgroups is congruent to $$1$$ modulo $$p$$.
 * 3) uses::Product formula

Proof
Given: A finite group $$G$$. A prime $$p$$, a $$p$$-subgroup $$A$$ of $$G$$.

To prove: The number of $$p$$-Sylow subgroups of $$G$$ containing $$A$$ is congruent to $$1$$ modulo $$p$$.

Proof: Let $$S = \operatorname{Syl}_p(G)$$ be the set of all $$p$$-Sylow subgroups of $$G$$. $$A$$ acts on the set $$S$$ by conjugation: every element of $$A$$ permutes the members of $$S$$ under conjugation.