## Intermediate Group Theory (2)

This is a continuation from the previous post. Let G act on set X, but now we assume that both G and X are finite. Since X is a disjoint union of transitive G-sets, and each transitive G-set is isomorphic to G/H for some subgroup H ≤ G, it follows that we have:

$|X| = \sum_i |G/H_i| = \sum_i \frac{|G|}{|H_i|}$.

Let’s consider the case where HiG. The corresponding G-sets are merely singleton sets since |G/Hi| = 1. These are the elements x of X for which gxx for all g in G. We call such elements fixed points of X and denote them XG.

Now we can modify our equation to obtain:

$|X| = |X_G| + \sum_i \frac{|G|}{|H_i|}$,

where each Hi is a proper subgroup of G. Let’s specialise even further.

Definition. Let p be prime. A finite group G≠{e} is called a p-group if its order is a power of p. A subgroup which is a p-group is also called a p-subgroup.

Plugging this into the above equation, if G is a p-group, then each |G|/|Hi| is a positive power of p and so is divisible by p. We thus obtain:

$|X| \equiv |X_G| \pmod p$.

This will be of great use to us.

## Sylow’s First Theorem

The three Sylow theorems are extremely powerful results regarding the structure of finite groups. Let’s start with the preliminary result.

Cauchy’s Theorem.  If |G| is divisible by prime p, then G has a subgroup of order p.

Proof. Use the following steps.

1. Let $X = \{(x_0, \dots, x_{p-1}) : x_i \in G, x_0 x_1 \ldots x_{p-1} = e\}$.
2. Let Z/p act on it by left-rotation: $m\cdot (x_0, x_1,\ldots, x_{p-1}) = (x_m, x_{m+1}, \ldots, x_{p-1}, x_0, \ldots, x_{m-1})$where m = 0, 1, …, p-1. [ Check that $x_m x_{m+1} \ldots x_{p-1} x_0 \ldots x_{m-1} = e$ even if G is not abelian. ]
3. Easy to count X : the first p-1 elements determine the final one uniquely, so $|X| = |G|^{p-1}$.
4. The set of fixed points XZ/p is precisely the set of (x, …, x) (x in G) such that xpe.
5. XZ/p is not empty since it contains (e, …, e).
6. Apply the above congruence: since Z/p is a p-group, |XZ/p| ≡ |X| ≡ 0 (mod p) so there’s a non-trivial element x of G such that xp = e. ♦

The first Sylow theorem is a more general version of Cauchy’s theorem, but its proof requires Cauchy’s theorem to kickstart it.

First Sylow Theorem. If |G| has order divisible by $p^{m+1}$ and H<G is a subgroup of order $p^m$ then H is contained in a subgroup K≤G of order $p^{m+1}$.

In short, by Cauchy’s theorem G has a subgroup H1 of order p. First Sylow theorem says this subgroup must be contained in another subgroup H1 < G of order p2. Applying this theorem repeatedly gives a sequence $H_1 < H_2 < H_3 \ldots \le G$ where each $|H_i| = p^i$.

Proof.

Case 1. If H is a normal subgroup of G, then G/H is a group of order divisible by p. Apply Cauchy’s theorem to obtain a subgroup G/H of order p, which corresponds to a group HK ≤ G of order × |H|. Done.

Case 2. If not, let’s try to “make it normal”. Hence, let $N(H) = \{g\in G: gHg^{-1}=H\}$. This is called the normaliser of H since it is the largest subgroup of G for which H is normal in. Thus, $H \triangleleft N(H)$ and it remains to show [N(H):H] is a multiple of p, after which we can replace G by N(H) and apply case 1. To show this,

1. consider the action of H on the set of left-cosets XG/H by left multiplication $(h, gH) \mapsto hgH$;
2. |G/H| is divisible by p by assumption, so |XH| is also divisible by p;
3. but XH is the set of gH for which hgHgH for all h in H; i.e. the set of gH for which gHg-1H; in short, XN(H)/H. ♦

## Second and Third Sylow Theorems

Let G be of order prm, where m is not divisible by prime p. Thanks to the first Sylow theorem, we know that there’s at least one subgroup of order pr. We shall call this a Sylow p-subgroup.

Second Sylow Theorem. Any two Sylow p-subgroups of G are conjugates, i.e. if $|H|=|P|=p^r$, then there is $g\in G$ such that $P = gHg^{-1}$.

More generally, if P is a Sylow p-subgroup of G, and H is a p-subgroup of G, then we can find $g\in G$, $gHg^{-1} \subseteq P$.

Proof.

Let H act on the set XG/P of left-cosets via left multiplication. Since P is Sylow p-subgroup, |X| is not a multiple of p. Thus, neither is |XH|. In particular, XH is not empty so contains some gP. Since this is fixed under action by H, we have hgPgP for all h in H, and so $g^{-1}Hg \subseteq P$ in particular. ♦

Third Sylow Theorem. Let $n_p$ be the number of Sylow p-subgroups of G. Then:

• $n_p$ divides |G| and
• $n_p \equiv 1 \pmod p$.

Proof.

First assertion: let P be any Sylow p-subgroup of G. Then all Sylow p-subgroups are of the form $gPg^{-1}$, $g\in G$. Repetition occurs iff

$xPx^{-1} = yPy^{-1} \iff y^{-1}xP(y^{-1}x)^{-1} = P \iff y^{-1}x \in N(P) \iff y\cdot N(P) = x\cdot N(P)$.

So the distinct Sylow p-subgroups correspond to the left cosets G/N(P), i.e. $n_p = |G|/|N(P)|$, so it divides |G|.

Second assertion: let P act on the set XG/P of left cosets (by left multiplication again). By the above congruence, we have:

$|X| \equiv |X_P| \pmod p$.

• Clearly LHS =  [GP].
• On the other hand, the set XP comprises of all gP such that xgPgP for all x in P. This holds iff $g^{-1}Pg = P$. But this precisely means that $g\in N(P)$. Hence $X_P = N(P)/P$ so RHS = [N(P): P].
• Since LHS is not divisible by p, dividing by RHS gives $[G:N(P)] \equiv 1 \pmod p$ which completes the proof. ♦
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