Congruence condition on Sylow numbers in terms of maximal Sylow intersection

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This article describes a congruence condition on an enumeration, or a count. It says that in a finite group and modulo prime number, the number of Sylow subgroups, i.e., the number of subgroups of prime power order whose index is relatively prime to the order satisfies a congruence condition.
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Statement

Let G be a finite group and p a prime dividing the order of G. Suppose the intersection of any two distinct p-Sylow subgroups has index at least p^r in both of them. Then, if n_p denotes the p-Sylow number (?) of G, i.e., the number of p-Sylow subgroups of G, we have:

n_p \equiv 1 \mod p^r.

In particular, n_p - 1 is a multiple of the index of a Maximal Sylow intersection (?).

Facts used

  1. Sylow subgroups exist
  2. Fundamental theorem of group actions
  3. Intersection of p-subgroup with Sylow subgroup equals intersection with normalizer: If P is a p-subgroup of G and Q is a p-Sylow subgroup of G, then P \cap Q = P \cap N_G(Q).

Proof

Given: A finite group G, a prime p dividing the order of G. The intersection of any two distinct p-Sylow subgroups has index at least p^r in both of them.

To prove: If n_p is the number of p-Sylow subgroups of G, then n_p \equiv 1 \mod p^r.

Proof: There exists at least one p-Sylow subgroup by fact (1). Let P be a p-Sylow subgroup. Let S be the set of all p-Sylow subgroups. P acts on S by conjugation: in other words, given g \in P and Q \in S, we define:

g \cdot Q = gQg^{-1}.

Suppose Q \in S and Q \ne P. Then, the size of the orbit of Q under the action of P is given by (using fact (2)):

\frac{|Q|}{\operatorname{Stab}_P(Q)}.

\operatorname{Stab}_P(Q) is the set of those elements of P that normalize Q, and is thus given by P \cap N_G(Q). By fact (3), this intersection equals P \cap Q. Hence, its index in Q, by assumption is a multiple of p^r, and thus, from the above expression, we get that the orbit of Q has size equal to a multiple of p^r.

Thus, the orbit of any element in S other than P has size a multiple of p^r. P itself, on the other hand, has an orbit of size one. Thus, the total number of members of S is congruent to 1 modulo p^r.