Fourth Sylow Theorem

Theorem
The number of Sylow $p$-subgroups of a finite group is congruent to $1 \pmod p$.

Proof
Let $G$ be a finite group such that $\order G = k p^n$ where $p \nmid k$ and $n > 0$.

Let $r$ be the number of Sylow $p$-subgroups of $G$.

We want to show that $r \equiv 1 \pmod p$.

Let $\mathbb S = \set {S \subseteq G: \card S = p^n}$, that is, the set of all of subsets of $G$ which have exactly $p^n$ elements.

From the reasoning in the First Sylow Theorem, we have:
 * $\size {\mathbb S} = \dbinom {p^n k} {p^n}$

Let $G$ act on $\mathbb S$ by the group action defined in Group Action on Sets with k Elements::
 * $\forall S \in \mathbb S: g \wedge S = g S = \set {x \in G: x = g s: s \in S}$

From Orbits of Group Action on Sets with Power of Prime Size:
 * there are exactly as many Sylow $p$-subgroups as there are orbits whose length is not divisible by $p$.

Also by Orbits of Group Action on Sets with Power of Prime Size:
 * all the terms in the Partition Equation are divisible by $k$, perhaps also divisible by $p$.

We can write the Partition Equation as:


 * $\size {\mathbb S} = \size {\Orb {S_1} } + \size {\Orb {S_2} } + \cdots + \size {\Orb {S_r} } + \size {\Orb {S_{r + 1} } } + \cdots + \size {\Orb {S_s} }$

where the first $r$ terms are the orbits containing the Sylow $p$-subgroups:
 * $\Stab {S_i}$

For each of these:
 * $\order G = \size {\Orb {S_i} } \times \size {\Stab {S_i} } = p^n \size {\Orb {S_i} }$

Thus:
 * $\size {\Orb {S_i} } = k$

for $1 \le i \le r$.

Each of the rest of the orbits are divisible by both $p$ and $k$, as we have seen.

So:
 * $\size {\mathbb S} = k r + m p k$

where:
 * the first term corresponds to the $r$ orbits containing the Sylow $p$-subgroups
 * the second term corresponds to all the rest of the orbits
 * $m$ is some unspecified integer.

That is, there exists some integer $m$ such that:
 * $\size {\mathbb S} = \dbinom {p^n k} {p^n} = k r + m p k$

Now this of course applies to the special case of the cyclic group $C_{p^n k}$.

In this case, there is exactly one subgroup for each divisor of $p^n k$.

In particular, there is exactly one subgroup of order $p^n$.

Hence, in this case:
 * $r = 1$

So we have an integer $m'$ such that $\dbinom {p^n k} {p^n} = k + m' p k$.

We can now equate these expressions:

and the proof is complete.

Also known as
Some sources call this the second Sylow theorem.

Others merge this result with what we call the Fifth Sylow Theorem and call it the third Sylow theorem.

Also see

 * Sylow Theorems