First Sylow Theorem

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

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

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

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.

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

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:

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

From Binomial Coefficient involving Power of Prime, $$\binom {p^n k} {p^n} \equiv k \left({\bmod\, p}\right)$$.

Thus, $$N \equiv k \left({\bmod\, p}\right)$$.


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

That is, $$g \wedge S$$ 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.

Each orbit is an equivalence class, and therefore 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 $$p \backslash N$$. But this can not be the case, because, as we have seen, $$N \equiv k \left({\bmod\, p}\right)$$.

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 $$\left|{\operatorname{Orb} \left({S}\right)}\right| = m: p \nmid m$$.

It follows from Group Action on Prime Power Order Subset that $$\operatorname{Stab} \left({S}\right) s = S$$, and thus $$\left|{\operatorname{Stab} \left({S}\right)}\right| = p^n$$.

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.