I read Introduction to Alexander Paulin's Abstract Algebra, and I could find 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 do not get is highlighted

Now remember that $ S $ is the disjoint union of the orbits of our action

of $ G $ sure $ S $.

(On the notation: $ Stab ( omega) $ is the stabilizing subgroup; $ Orb (omega) $ is the orbit; and HCF means the highest common factor):

Sylow's theorem. Let $ (G, *) $ to be a finite group such as $ p ^ n $ Split $ | G | $, or $ p $ is first.

Then there is a subgroup of order $ p ^ n $.

Evidence. Suppose that $ | G | = p ^ nm $, or $ m = p ^ ru $ with $ HCF (p, u) = $ 1. Our central strategy

is to consider an intelligently chosen group action of G and to prove one of the subgroups of stabilizers

to the size $ p ^ n $. We will need to heavily exploit the orbit stabilizer theorem.

Let $ S $ be the set of all subsets of $ G $ size $ p ^ n $. An element of $ S $ is an unordered $ n $-tuple

of distinct elements in $ G $. There is a natural action of $ G $ sure $ S $ by term term composition

on the left.

Let $ ∈ S $. If we correct an order $ ω = {ω_1, · · ·, _ {p ^ n}} S $then $ g (ω): = {g * _1, · · ·, g * _ {p ^ n}} $.

• We claim first that $ | Stab (ω) | ≤ p ^ n $. To see this define the function

$$ f: Stab (ω) → ω $$

$$ g → g * _1 $$

By the property of cancellation for groups it is an injective card. Therefore $ | Stab (ω) | ≤ | ω | = p ^ n $.

• Observe this

$$ | S | = pmatrix {p ^ nm \ p ^ n} = frac {(p ^ nm)!} {p ^ n! (p ^ mp ^ n)!} = prod_ {j = 0} ^ {p ^ n-1} frac {p ^ nm-j} {p ^ nj} = m prod_ {j = 1} ^ { p ^ n-1} frac {p ^ mj} {p ^ nj} $$

Note that if $ 1 ≤ j ≤ p ^ n – $ 1 then j is divisible by $ p $ at most $ n – $ 1 time. It means

this $ p ^ nm – j $ and $ p ^ n – j $ have the same number of $ p $ factors, namely the number of $ p $

factor of $ j. It means that

$$ m prod_ {j = 1} ^ {p ^ n-1} frac {p ^ m-j} {p ^ n-j} $$

do not have $ p $ factors. Therefore $ | S | = p ^ rv $, or $ HCF (p, v) = $ 1.

## Now remember that $ S $ is the disjoint union of the orbits of our action of $ G $ sure $ S $.

Therefore

there must be a $ ∈ S $ such as $ | Orb (ω) | = p ^ st $, or $ s ≤ r $ and $ HCF (p, t) = $ 1.

By

the orbiting 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 can deduce $ frac {u} {t} N $. Therefore $ | Stab (ω) | ≥ p ^ n $.

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

So I do not understand this line:

Now remember that $ S $ is the disjoint union of the orbits of our action

of $ G $ sure $ 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 counterclockwise and clockwise rotation; $ c $,$ d $,and $ f $ like reflections. Let $ p = $ 3, $ n = $ 1, $ m = $ 2.

then $ S $ would be the set of all subsets of $ Sym_3 $ size $ 3 $. Picking $ 3 $ elements of $ 6 $ at $ pmatrix {6 \ 3} = $ 20 choice, so $ S $ at the size of $ 20 $to know

$$ 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, what disjointed orbits formed this $ S $?