First Sylow Theorem/Proof 2

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

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

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

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


 * $(1): \quad \size {\mathbb S} = \size {\Orb {S_1} } + \card {\Orb {S_2} } + \ldots + \size {\Orb {S_r} }$

Thus each $\Orb {S_i}$ is the orbit under $*$ of some $S_i$ whose order is $p^n$.

By the Orbit-Stabilizer Theorem:
 * $(2): \quad \order {\Orb {S_i} } = \dfrac {\order G} {\order {\Stab {S_i} } }$ for all $i \in \set {1, 2, \ldots, n}$

where $\Stab {S_i}$ is the stabilizer of $S_i$ under $*$.

Let $s \in S_i$ and $x \in \Stab {S_i}$.

Then $s x \in S_i$ because $S_i x = S_i$.

Let $s$ be fixed while $x$ runs over $\Stab {S_i}$.

Then:
 * $\forall s \in S_i: s \Stab {S_i} \subseteq S_i$

Therefore $S_i$ is the union of all the left cosets $s \Stab {S_i}$.

These left cosets may not all be distinct.

Let $r_i$ be the number of distinct left cosets of $s \Stab {S_i}$.

Then:
 * $\card {S_i} = r_i \card {\Stab {S_i} }$

because:
 * distinct left cosets are disjoint
 * all have the same cardinality $\card {\Stab {S_i} }$.

Thus:
 * $(3): \quad \card {\Stab {S_i} } \divides \card {S_i} = p^n$

Because the only divisors of $p^n$ are smaller powers of $p$:


 * $\card {\Stab {S_i} } = p^{n_i}$

for some $n_i \le n$, for $i = 1, 2, \ldots, r$.

We can write:
 * $g = \order G = k p^n$

where $p$ does not divide $k$.

From $(2)$ and $(3)$:


 * $\card {S_i} = k p^{d_i}$

where $d_i = n - n_i$ for $i = 1, 2, \ldots, r$.

From Cardinality of Set of Subsets:
 * $\card {\Bbb S} = \dbinom {k p^n} {p_n}$

We have from Binomial Coefficient involving Power of Prime that:
 * $\dbinom {k p^n} {p_n} \equiv k \pmod p$

But $p$ does not divide $k$.

Hence $p$ does not divide $\card {\Bbb S}$.

Hence by $(1)$ and $(4)$:
 * $p \nmid k \paren {p^{d_1} + p^{d_2} + \cdots p^{d_r} }$

If $d_i > 0$ for all $i = 1, 2, \ldots, r$, then $p$ would divide this integer.

So there must be some $i$ such that $d_i = 0$.

Then, for this $i$:
 * $n_i = n$

That is:
 * $\card {\Stab {S_i} } = p^n$

Thus $\Stab {S_i}$ is the subgroup of $G$ of order $p_n$ that was asserted.