Fourth Sylow Theorem

Theorem
The number of Sylow $p$-subgroups of a finite group is congruent to $$1 \left({\bmod\, p}\right)$$.

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.

Proof
Let $$G$$ be a finite group such that $$\left|{G}\right| = k p^n$$ where $$p \nmid k$$ and $$n > 0$$.

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

We want to show that $$r \equiv 1 \left({\bmod\, p}\right)$$.


 * Let $$\mathbb{S} = \left\{{S \subseteq G: \left|{S}\right| = p^n}\right\}$$, 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 $$\left|{\mathbb{S}}\right| = \binom {p^n k} {p^n}$$.

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

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 $$k$$. Some of them may also be divisible by $$p$$.

We can write the Partition Equation as:

$$ \left|{\mathbb{S}}\right| = \left|{\operatorname{Orb} \left({S_1}\right)}\right| + \left|{\operatorname{Orb} \left({S_2}\right)}\right| + \cdots + \left|{\operatorname{Orb} \left({S_r}\right)}\right| + \left|{\operatorname{Orb} \left({S_{r+1}}\right)}\right| + \cdots + \left|{\operatorname{Orb} \left({S_s}\right)}\right|$$

where the first $$r$$ terms are the orbits containing the Sylow $p$-subgroups: $$\operatorname{Stab} \left({S_i}\right)$$.

For each of these, $$\left|{G}\right| = \left|{\operatorname{Orb} \left({S_i}\right)}\right| \times \left|{\operatorname{Stab} \left({S_i}\right)}\right| = p^n \left|{\operatorname{Orb} \left({S_i}\right)}\right|$$.

Thus $$\left|{\operatorname{Orb} \left({S_i}\right)}\right| = k$$ 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 $$\left|{\mathbb{S}}\right| = k r + m p k$$, where the first term corresponds to the $$r$$ orbits containing the Sylow $$p$$-subgroups, and the second term to all the rest of the orbits, and $$m$$ is some unspecified integer.

That is, there exists some integer $$m$$ such that $$\left|{\mathbb{S}}\right| = \binom {p^n k} {p^n} = k r + m p k$$.

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 $$\binom {p^n k} {p^n} = k + m' p k$$.

We can now equate these expressions:

$$ $$ $$ $$ $$

and the proof is complete.