11-1100-Pgadey-Lect6

Theorem

Every $G$-set is a disjoint union of "transitive $G$-sets"

Theorem

If <<X>> is a transitive <<G>>-set and <<x \in X>> then <<X \simeq G/Stab(x)>> where the isomorphism an isomorphism of <<G>>-sets.

Transitive <<G>>-set

A <<G>>-set <<X>> is transitive is <<\forall_{x,y \in X} \exits_{g \in G}\ st.\ gx = y>>.

Stabilizer of a point

We write <<Stab(x) = \{g \in G : gx = x\}>> for the stabilizer subgroup of $x$.

Proof We define an equivalence relation <<x \sim y \iff \exists_{g \in G} gx = y>>. This relation is reflexive since <<x = ex>> and thus <<x \sim x>>. This relation is symmetric since <<y = gx>> implies <<g^{-1}y = x>>. This relation is transitive, since if <<x = gy>> and <<y = hz>> then <<x = ghz>>. It follows that << X = \coprod_{i \in I} Gx_{i} >> where <<Gx_i>> denote the orbit of a point <<x_i>>.

We then claim that <<Gx_i>> is a transitive <<G>>-set. [Dror: "[This fact] is too easy."]

We show that <<Gx>> is isomorphic to <<G / Stab(x)>> as a <<G>>-set.

We produce two morphism <<\Psi : Gx \rightarrow G/Stab(x)>> and <<\Phi : G/Stab(x) \rightarrow Gx>>.

To define <<\Psi>> there is only one thing we can do. We have <<y \in Gx \Rightarrow y = gx>> and then we define <<\Psi(y) = g Stab(x)>>. We check that this map is well defined. If <<y = gx = g'x>> then <<g^{-1}g'x = x>> and hence <<g^{-1}g \in Stab(x)>>. It follows that <<gStab(x) = g'Stab(x)>>. Thus <<\Psi>> is well defined.

To define <<\Phi>> we take <<gStab(x) \in G/Stab(x)>> and define <<\Phi(gStab(x)) = gx>>. We show that this map is well defined. If <<gStab(x) = g'Stab(x)>> then <<g^{-1}g' \in Stab(x)>> and hence <<g^{-1}g'x = x>>. It follows that <<gx = g'x>> and hence <<\Phi>> is well defined.

We need to check that <<\Psi>> and <<\Phi>> are mutually inverse and <<G>>-set morphisms. We quickly check that <<\Phi>> is a <<G>>-set morphism. If <<y = gx>> and <<g_1 \in G>> then <<g_1\Psi(y) = g_1(gStab(x)) = (g_1g)Stab(x)>>. Similarly, <<\Psi(g_1y) = g'Stab(x) = (g_1g)Stab(x)>>. The last inequality follows since we can take any <<g'>> such that <<g'y = g_1y>>. Why not take <<g' = g_1g>> -- since we know that works.

Theorem (Orbit-Stabilizer)

If <<|X| < \infty>> and <<X = \coprod_{i \in I} Gx_i>> then <<|X| = \sum_{i} \frac{|G|}{Stab(x_i)}>>.

This is just a rewriting of the theorem above.

<

>-Group

A <

>-group is a group <<G>> with <<|G| = p^k>> for some <<k>>.

<<G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}>>

The last group <> is the famous unit quaternions -- They need a better description here.

Theorem

Any <

>-group has a non-trivial centre.

Let <<G>> act on itself by conjugation. Decompose <<G = \coprod Gx_i>>. Then, << |G| = \sum_{|Gx_i| = 1} 1 + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} >> Observe that <<|Gx_i|| = 1>> iff <<x_i \in Z(G)>>. It follows that << |G| = |Z(G)| + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} >>

The formula above is called "the class formula". We have that <<|G| / |Stab(x)| = p^k>> for some << 1 < k>> since <<Stab(x)>> is a subgroup. It follows that <<|G| \equiv 0\ \mod\ p>> and <<\sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x_i)|} \equiv 0\ \mod\ p>>. It follows that <<|Z(G)| \equiv 0\ \mod\ p>>. Since <<e \in Z(G)>> we have <<1 \leq |Z(G)|>> and thus <

>. SYLOW A prove a brief technical lemma, for fun, since we could deduce it from more high powered machinery which we don't have yet.

Cauchy's Lemma

If <<A>> is an abelian group and <

> divides <<|A|>>, then there is an element of order <

> in <<A>>.

Proof. Pick <<x \in A>>. If <

> divides the order of <<x>> then we have <<x^{np} = e>> for some <<n>>. It follows that <<(x^n)^p = e>>. We then have that the order of <<x^n>> is <

>. If <

> does not divide the order of <

>, then consider <<A / <x> >>. Since <<A>> is abelian, << <x> >> is a normal subgroup. We have that <

> divides <<|A/<x>|>>, and <<|A / <x>| < |A|>>. We then induct. Let << y<x> >> have order <

>, that is << (y<x>)^p = <x> >>. We then have that << y^p = x^k >> for some << k >>. We write << |<y>| = np + r >> where << 0 \leq p < r >>. We then have << e = y^{|<y>|} = y^{np + r} = x^{nkp}y^r \Rightarrow y^r \in <x> >>. It follows that <<(y<x>)^r = <x> >> contradicting the assumption that the order of << y<x> >> is << p >>.

Sylow set

If <<|G| = p^k m>> for <<m \not\equiv 0\ \mod\ p>> then <<Syl_p(G) = \{P \leq G : |P| = p^k>>.

Sylow I

<<Syl_p(G) \neq \emptyset>>

We proceed by induction on the oder of $p$. Assume the claim holds for all groups of order less than $|G|$. [Dror: "Stare at the class equation.] Since <<|G| \equiv 0\ \mod\ p>> we have either:

• <<|G| \equiv 0\ \mod\ p>> and <<\sum |G|/|Stab(x_i)| \equiv 0\ \mod\ p>>.
• <<|G| \equiv 0\ \not\mod\ p>> and <<\sum |G|/|Stab(x_i)| \not\equiv 0\ \mod\ p>>.

If <<|Z(G)| \not\equiv 0\ \mod p>> then there exists <<x_i>> such that <<|G|/|Stab(x_i)| \not\equiv 0\ \mod\ p>>. Thus <<p^k>> divides <<|Stab(x_i)|>>. We have that <<|Stab(x_i)| < |G|>> [Why happens here?] We then have that <<p^k \leq Stab(x_i) < |G|>> and by induction there is <<|P| = p^k>> such that <

>. It follows <

>. We've obtained the Sylow <

>-subgroup. WIf <<|Z(G)| \equiv 0\ \mod p>> then by Cauchy's Lemma, there is <<x \in Z(G)>> with <<|<x>| = p>>. Consider the group << G / <x> >>. By the induction hypothesis there is << P' \leq G/<x> >> where <<|P'| = p^{k-1}>>. Then, there is the canonical projection << \pi : G \rightarrow G/<x> >>. By the fourth isomorphism theory << P = \pi^{-1}(P') \leq G >> and << |\pi^{-1}(P')| = p(p^{k-1}) = p^k >>.

Sylow 2

Every Sylow <

>-subgroup of <<G> is conjugate. Moreover, every <

>-subgroup is contained in a Sylow <

>-subgroup.

Sylow 3

Let <<n_p(G) = |Syl_p(G)|>>. We have <<n_p \equiv 0\ \mod\ |G|>> and <<n_p \equiv 1\ \mod\ p>>.

A Nearly Tautological Lemma

If <

> and <<H \lea N(P)>> is a <

>-group, then <<H \leq P>>.

If <<x \in G>> has <<|<x>| = p^k>> and <<x \in N(P)>> then <<x \in P>>.

[Dror: "This lemma is nearly tautological but it is only nearly tautological once you understand that it is nearly tautological." Parker: "A tautology?"] We show the first statement. We have that <<|P / P \cap H| = p^k>> since <

> is a <

>-group. We then know that <<PH / H \simeq P / P \cap H>> by the second isomorphism theorem. It foolows that <<|PH| = p^{k'}>>. But since <

> is maximal, we have <

> and thus <<H \subseteq P>>. The first statement implies the second by taking <<H = <x> >>. GROUPS OF ORDER 15 If <<|G| = 15>> then <<n_3 \equiv 0\ \mod\ 15>> and <<n_3 \equiv 1\ \mod\ 3>>. These imply <<n_3 = 1>>. Moreover, <<n_5 \equiv 0\ \mod\ 15>> and <<n_5 \equiv 1\ \mod\ 5>>. These imply <<n_5 = 1>>. Thus we have <<P_3>> a normal <<3>>-subgroup. Moreover, we have <<P_5>> a normal <<5>>-subgroup. This tells us a lot about the group.