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 H_i = G. The corresponding G-sets are merely singleton sets since |G/H_i| = 1. These are the elements x of X for which gx = x for all g in G. We call such elements fixed points of X and denote them X_G.

Now we can modify our equation to obtain:

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

where each H_i 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|/|H_i| 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.

Let $X = \{(x_0, \dots, x_{p-1}) : x_i \in G, x_0 x_1 \ldots x_{p-1} = e\}$ .
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. ]
Easy to count X : the first p-1 elements determine the final one uniquely, so $|X| = |G|^{p-1}$ .
The set of fixed points X_Z/p is precisely the set of (x, …, x) (x in G) such that x^p = e.
X_Z/p is not empty since it contains (e, …, e).
Apply the above congruence: since Z/p is a p-group, |X_Z/p| ≡ |X| ≡ 0 (mod p) so there’s a non-trivial element x of G such that x^p = 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 H < K ≤ G of order p × |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,

consider the action of H on the set of left-cosets X = G/H by left multiplication $(h, gH) \mapsto hgH$ ;
|G/H| is divisible by p by assumption, so |X_H| is also divisible by p;
but X_H is the set of gH for which hgH = gH for all h in H; i.e. the set of gH for which gHg^-1 = H; in short, X_H= N(H)/H. ♦

Second and Third Sylow Theorems

Let G be of order p^rm, 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 p^r. 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 X = G/P of left-cosets via left multiplication. Since P is Sylow p-subgroup, |X| is not a multiple of p. Thus, neither is |X_H|. In particular, X_H is not empty so contains some gP. Since this is fixed under action by H, we have hgP = gP 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 X = G/P of left cosets (by left multiplication again). By the above congruence, we have:

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

Clearly LHS = [G: P].
On the other hand, the set X_P comprises of all gP such that xgP = gP 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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