# Proof of Sylow's Theorem

I'm reading Alexander Paulin's Introduction to Abstract Algebra, and could not figure out a line in the proof of Sylow's Theorem.

It's on the 24th and 25th page of the notes, quoted below, where the line I don't get is highlighted

Now recall that $$S$$ is the disjoint union of the orbits of our action of $$G$$ on $$S$$.

(On notation: $$Stab(\omega)$$ is the stabilizer subgroup; $$Orb(\omega)$$ is the orbit; and HCF means Hightest Common Factor) :

Sylow’s Theorem. Let $$(G, ∗)$$ be a finite group such that $$p^n$$ divides $$|G|$$, where $$p$$ is prime. Then there exists a subgroup of order $$p^n$$.

Proof. Assume that $$|G| = p^nm$$, where $$m = p^ru$$ with $$HCF(p, u) = 1$$. Our central strategy is to consider a cleverly chosen group action of G and prove one of the stabilizer subgroups has size $$p^n$$. We’ll need to heavily exploit the orbit-stabilizer theorem.

Let $$S$$ be the set of all subsets of $$G$$ of size $$p^n$$. An element of $$S$$ is an unordered $$n$$-tuple of distinct elements in $$G$$. There is a natural action of $$G$$ on $$S$$ by term-by-term composition on the left.

Let $$ω ∈ S$$. If we fix an ordering $$ω = {ω_1, · · · , ω_{p^n}} ∈ S$$, then $$g(ω) := {g∗ω_1, · · · , g∗ω_{p^n}}$$.

• We first claim that $$|Stab(ω)| ≤ p^n$$. To see this define the function $$f : Stab(ω) → ω$$ $$g → g ∗ ω_1$$ By the cancellation property for groups this is an injective map. Hence $$|Stab(ω)| ≤ |ω| = p^n$$.

• Observe that $$|S| = \pmatrix{p^nm\\p^n} = \frac{(p^nm)!}{p^n!(p^m-p^n)!} = \prod_{j=0}^{p^n-1}\frac{p^nm-j}{p^n-j}=m\prod_{j=1}^{p^n-1}\frac{p^m-j}{p^n-j}$$

Observe that if $$1 ≤ j ≤ p^n−1$$ then j is divisible by $$p$$ at most $$n−1$$ times. This means that $$p^nm − j$$ and $$p^n − j$$ have the same number of $$p$$ factors, namely the number of $$p$$ factor of $$j$$. This means that $$m\prod_{j=1}^{p^n-1}\frac{p^m-j}{p^n-j}$$

has no $$p$$ factors. Hence $$|S| = p^rv$$, where $$HCF(p, v) = 1$$.

## Now recall that $$S$$ is the disjoint union of the orbits of our action of $$G$$ on $$S$$.

Hence there must be an $$ω ∈ S$$ such that $$|Orb(ω)| = p^st$$, where $$s ≤ r$$ and $$HCF(p, t) = 1$$.

By the orbit-stabilizer theorem we know that $$|Stab(ω)| = p^{n+r−s} \frac{u}{t}$$.

Because $$|Stab(ω)| ∈ N$$ and $$u$$ and $$t$$ are coprime to $$p$$, we deduce that $$\frac{u}{t} ∈ N$$. Hence $$|Stab(ω)| ≥ p^n$$.

• For this choice of ω ∈ S, Stab(ω) is thus a subgroup of size pn.

So I don't understand this line:

Now recall that $$S$$ is the disjoint union of the orbits of our action of $$G$$ on $$S$$.

What does it mean? for example, if take $$G$$ as $$Sym_3=\{e,a,b,c,d,f\}$$, the transformers of a triangle, where $$a$$, $$b$$ are the anti-clockwise and clockwise rotation; $$c$$,$$d$$,and $$f$$ as the reflections. Let $$p=3$$, $$n=1$$, $$m=2$$.

Then $$S$$ would be the set of all subsets of $$Sym_3$$ of size $$3$$. Picking $$3$$ elements from $$6$$ has $$\pmatrix{6\\3} = 20$$ choices, so $$S$$ has the size of $$20$$, namely $$S=\{\{e,a,b\},\{e,a,c\},\{e,a,d\},\{e,a,f\},\{e,b,c\},\{e,b,d\},\{e,b,f\},\{e,c,d\},\{e,c,f\},\{e,d,f\}, \{a,b,c\},\{a,b,d\},\{a,b,f\},\{a,c,d\},\{a,c,f\},\{a,d,f\}, \{b,c,d\},\{b,c,f\},\{b,d,f\},\{c,d,f\}$$

So which disjointed orbits formed this $$S$$?

If $$S$$ is any set and $$G$$ is a group that acts on $$S$$, remember that the orbit of an element $$s \in S$$ is the set $$Gs = \{g.s \mid g \in G\}$$. You should prove the following: if $$s$$ and $$t$$ are elements of $$S$$, then either $$Gs = Gt$$, or $$Gs \cap Gt = \varnothing$$. That is, orbits are either identical or disjoint.
The claim follows: $$S$$ is the disjoint union of the elements of the set $$\{Gs \mid s \in S\}$$. Like the textbook, I've pulled a fast one: I only really care about the $$Gs$$ when they are different.