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 $S$?
Best Answer
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.