First Sylow Theorem

Theorem
Let $p$ be a prime number, and let $G$ be a group such that $\left\vert{G}\right\vert = k p^n$ where $p \nmid k$.

Then $G$ has at least one Sylow $p$-subgroup.

Proof
Let $\left\vert{G}\right\vert = k p^n$ such that $p \nmid k$.

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

Let $N = \left\vert{\mathbb S}\right\vert$.

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:


 * $\displaystyle N = \binom {p^n k} {p^n} = \frac {\left({p^n k}\right) \left({p^n k - 1}\right) \cdots \left({p^n k - i}\right) \cdots \left({p^n k - p^n + 1}\right)} {\left({p^n}\right) \left({p^n - 1}\right) \cdots \left({p^n - i}\right) \cdots \left({1}\right)}$

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

Thus, $N \equiv k \pmod p$.

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

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 Orbit is Equivalence Class, the orbits partition $\mathbb S$.

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

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

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 $S \in \left\{{S_1, S_2, \ldots, S_r}\right\}$ be such that $\left\vert{\operatorname{Orb} \left({S}\right)}\right\vert = m: p \nmid m$.

It follows from Group Action on Prime Power Order Subset that:
 * $\operatorname{Stab} \left({S}\right) s = S$

and so:
 * $\left\vert{\operatorname{Stab} \left({S}\right)}\right\vert = p^n$

From Stabilizer is Subgroup:
 * $\operatorname{Stab} \left({S}\right) \le G$

Thus $\operatorname{Stab} \left({S}\right)$ is the subgroup of $G$ with $p^n$ elements of which we wanted to prove the existence.

Note
By Orbits of Group Action on Sets with Power of Prime Size, it was clear that $k \mathop \backslash \left|{\operatorname{Orb} \left({S}\right)}\right|$.

However here, since it is established that $\left\vert{\operatorname{Stab} \left({S}\right)}\right\vert = p^n$, and by Orbit-Stabilizer Theorem, we also have $k = \left\vert{\operatorname{Orb} \left({S}\right)}\right\vert$.