First Sylow Theorem/Proof 1

Proof
Let $\order G = k p^n$ such that $p \nmid k$.

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

Let $N = \order {\mathbb S}$.

Now $N$ is the number of ways $p^n$ elements can be chosen from a set containing $p^n k$ elements. From Cardinality of Set of Subsets, this is given by:


 * $N = \dbinom {p^n k} {p^n} = \dfrac {\paren {p^n k} \paren {p^n k - 1} \cdots \paren {p^n k - i} \cdots \paren {p^n k - p^n + 1} } {\paren {p^n} \paren {p^n - 1} \cdots \paren {p^n - i} \cdots \paren 1}$

From Binomial Coefficient involving Power of Prime:
 * $\dbinom {p^n k} {p^n} \equiv k \pmod p$

Thus:
 * $N \equiv k \pmod p$

Now let $G$ act on $\mathbb S$ by the rule:
 * $\forall S \in \mathbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$

That is, $g * S$ is the left coset of $S$ by $g$.

From Group Action on Sets with k Elements, this is a group action.

Now, let $\mathbb S$ have $r$ orbits under this action.

From Set of Orbits forms Partition, the orbits partition $\mathbb S$.

Let these orbits be represented by $\set {S_1, S_2, \ldots, S_r}$, so that:

If each orbit had length divisible by $p$, then $p \divides N$.

But this can not be the case, because, as we have seen:
 * $N \equiv k \pmod p$

So at least one orbit has length which is not divisible by $p$.

Let $S \in \set {S_1, S_2, \ldots, S_r}$ be such that $\size {\Orb S)} = m: p \nmid m$.

Let $s \in S$.

It follows from Group Action on Prime Power Order Subset that:
 * $\Stab S s = S$

and so:
 * $\size {\Stab S} = \size S = p^n$

From Stabilizer is Subgroup:
 * $\Stab S \le G$

Thus $\Stab S$ is the subgroup of $G$ with $p^n$ elements of which we wanted to prove the existence.