11-1100-Pgadey-Lect6

Theorem

Every [math]\displaystyle{ G }[/math]-set is a disjoint union of "transitive [math]\displaystyle{ G }[/math]-sets"

Theorem

If [math]\displaystyle{ X }[/math] is a transitive [math]\displaystyle{ G }[/math]-set and [math]\displaystyle{ x \in X }[/math] then [math]\displaystyle{ X \simeq G/Stab(x) }[/math] where the isomorphism an isomorphism of [math]\displaystyle{ G }[/math]-sets.

Transitive [math]\displaystyle{ G }[/math]-set

A [math]\displaystyle{ G }[/math]-set [math]\displaystyle{ X }[/math] is transitive is [math]\displaystyle{ \forall_{x,y \in X} \exists_{g \in G}\ st.\ gx = y }[/math].

Stabilizer of a point

We write [math]\displaystyle{ Stab(x) = \{g \in G : gx = x\} }[/math] for the stabilizer subgroup of $x$.

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

We then claim that [math]\displaystyle{ Gx_i }[/math] is a transitive [math]\displaystyle{ G }[/math]-set. [Dror: "[This fact] is too easy."]

We show that [math]\displaystyle{ Gx }[/math] is isomorphic to [math]\displaystyle{ G / Stab(x) }[/math] as a [math]\displaystyle{ G }[/math]-set.

We produce two morphism [math]\displaystyle{ \Psi : Gx \rightarrow G/Stab(x) }[/math] and [math]\displaystyle{ \Phi : G/Stab(x) \rightarrow Gx }[/math].

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

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

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

Theorem (Orbit-Stabilizer)

If [math]\displaystyle{ |X| \lt \infty }[/math] and [math]\displaystyle{ X = \coprod_{i \in I} Gx_i }[/math] then [math]\displaystyle{ |X| = \sum_{i} \frac{|G|}{Stab(x_i)} }[/math].

This is just a rewriting of the theorem above.

[math]\displaystyle{ p }[/math]-Group

A [math]\displaystyle{ p }[/math]-group is a group [math]\displaystyle{ G }[/math] with [math]\displaystyle{ |G| = p^k }[/math] for some [math]\displaystyle{ k }[/math].

[math]\displaystyle{ 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\} }[/math]

The last group [math]\displaystyle{ Q }[/math] is the famous unit quaternions -- They need a better description here.

Theorem

Any [math]\displaystyle{ p }[/math]-group has a non-trivial centre.

Let [math]\displaystyle{ G }[/math] act on itself by conjugation. Decompose [math]\displaystyle{ G = \coprod Gx_i }[/math]. Then, [math]\displaystyle{ |G| = \sum_{|Gx_i| = 1} 1 + \sum_{|Gx_i| \gt 1} \frac{|G|}{|Stab(x)|} }[/math] Observe that [math]\displaystyle{ |Gx_i|| = 1 }[/math] iff [math]\displaystyle{ x_i \in Z(G) }[/math]. It follows that [math]\displaystyle{ |G| = |Z(G)| + \sum_{|Gx_i| \gt 1} \frac{|G|}{|Stab(x)|} }[/math] The formula above is called "the class formula". We have that [math]\displaystyle{ |G| / |Stab(x)| = p^k }[/math] for some [math]\displaystyle{ 1 \lt k }[/math] since [math]\displaystyle{ Stab(x) }[/math] is a subgroup. It follows that [math]\displaystyle{ |G| \equiv 0\ \mod\ p }[/math] and [math]\displaystyle{ \sum_{|Gx_i| \gt 1} \frac{|G|}{|Stab(x_i)|} \equiv 0\ \mod\ p }[/math]. It follows that [math]\displaystyle{ |Z(G)| \equiv 0\ \mod\ p }[/math]. Since [math]\displaystyle{ e \in Z(G) }[/math] we have [math]\displaystyle{ 1 \leq |Z(G)| }[/math] and thus [math]\displaystyle{ p \leq |Z(G)| }[/math].

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 [math]\displaystyle{ A }[/math] is an abelian group and [math]\displaystyle{ p }[/math] divides [math]\displaystyle{ |A| }[/math], then there is an element of order [math]\displaystyle{ p }[/math] in [math]\displaystyle{ A }[/math].

Proof. Pick [math]\displaystyle{ x \in A }[/math]. If [math]\displaystyle{ p }[/math] divides the order of [math]\displaystyle{ x }[/math] then we have [math]\displaystyle{ x^{np} = e }[/math] for some [math]\displaystyle{ n }[/math]. It follows that [math]\displaystyle{ (x^n)^p = e }[/math]. We then have that the order of [math]\displaystyle{ x^n }[/math] is [math]\displaystyle{ p }[/math]. If [math]\displaystyle{ p }[/math] does not divide the order of [math]\displaystyle{ p }[/math], then consider [math]\displaystyle{ A / \lt x\gt }[/math]. Since [math]\displaystyle{ A }[/math] is abelian, [math]\displaystyle{ \lt x\gt }[/math] is a normal subgroup. We have that [math]\displaystyle{ p }[/math] divides [math]\displaystyle{ |A/\lt x\gt | }[/math], and [math]\displaystyle{ |A / \lt x\gt | \lt |A| }[/math]. We then induct. Let [math]\displaystyle{ y\lt x\gt }[/math] have order [math]\displaystyle{ p }[/math], that is [math]\displaystyle{ (y\lt x\gt )^p = \lt x\gt }[/math]. We then have that [math]\displaystyle{ y^p = x^k }[/math] for some [math]\displaystyle{ k }[/math]. We write [math]\displaystyle{ |\lt y\gt | = np + r }[/math] where [math]\displaystyle{ 0 \leq p \lt r }[/math]. We then have [math]\displaystyle{ e = y^{|\lt y\gt |} = y^{np + r} = x^{nkp}y^r \Rightarrow y^r \in \lt x\gt }[/math]. It follows that [math]\displaystyle{ (y\lt x\gt )^r = \lt x\gt }[/math] contradicting the assumption that the order of [math]\displaystyle{ y\lt x\gt }[/math] is [math]\displaystyle{ p }[/math].

Sylow set

If [math]\displaystyle{ |G| = p^k m }[/math] for [math]\displaystyle{ m \not\equiv 0\ \mod\ p }[/math] then [math]\displaystyle{ Syl_p(G) = \{P \leq G : |P| = p^k }[/math].

Sylow I

[math]\displaystyle{ Syl_p(G) \neq \emptyset }[/math]

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 [math]\displaystyle{ |G| \equiv 0\ \mod\ p }[/math] we have either:

• [math]\displaystyle{ |G| \equiv 0\ \mod\ p }[/math] and [math]\displaystyle{ \sum |G|/|Stab(x_i)| \equiv 0\ \mod\ p }[/math].
• [math]\displaystyle{ |G| \equiv 0\ \not\mod\ p }[/math] and [math]\displaystyle{ \sum |G|/|Stab(x_i)| \not\equiv 0\ \mod\ p }[/math].

If [math]\displaystyle{ |Z(G)| \not\equiv 0\ \mod p }[/math] then there exists [math]\displaystyle{ x_i }[/math] such that [math]\displaystyle{ |G|/|Stab(x_i)| \not\equiv 0\ \mod\ p }[/math]. Thus [math]\displaystyle{ p^k }[/math] divides [math]\displaystyle{ |Stab(x_i)| }[/math]. We have that [math]\displaystyle{ |Stab(x_i)| \lt |G| }[/math] [Why happens here?] We then have that [math]\displaystyle{ p^k \leq Stab(x_i) \lt |G| }[/math] and by induction there is [math]\displaystyle{ |P| = p^k }[/math] such that [math]\displaystyle{ P \leq Stab(x_i) }[/math]. It follows [math]\displaystyle{ P \leq Stab(x_i) \leq G }[/math]. We've obtained the Sylow [math]\displaystyle{ p }[/math]-subgroup.

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

Sylow 2

Every Sylow [math]\displaystyle{ p }[/math]-subgroup of [math]\displaystyle{ G\gt is conjugate. Moreover, every \lt math\gt p }[/math]-subgroup is contained in a Sylow [math]\displaystyle{ p }[/math]-subgroup.

Sylow 3

Let [math]\displaystyle{ n_p(G) = |Syl_p(G)| }[/math]. We have [math]\displaystyle{ n_p \equiv 0\ \mod\ |G| }[/math] and [math]\displaystyle{ n_p \equiv 1\ \mod\ p }[/math].

A Nearly Tautological Lemma

If [math]\displaystyle{ P \in Syl_p(G) }[/math] and [math]\displaystyle{ H \lea N(P) }[/math] is a [math]\displaystyle{ p }[/math]-group, then [math]\displaystyle{ H \leq P }[/math].

If [math]\displaystyle{ x \in G }[/math] has [math]\displaystyle{ |\lt x\gt | = p^k }[/math] and [math]\displaystyle{ x \in N(P) }[/math] then [math]\displaystyle{ x \in P }[/math].

[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 [math]\displaystyle{ |P / P \cap H| = p^k }[/math] since [math]\displaystyle{ P }[/math] is a [math]\displaystyle{ p }[/math]-group. We then know that [math]\displaystyle{ PH / H \simeq P / P \cap H }[/math] by the second isomorphism theorem. It foolows that [math]\displaystyle{ |PH| = p^{k'} }[/math]. But since [math]\displaystyle{ P }[/math] is maximal, we have [math]\displaystyle{ P = PH }[/math] and thus [math]\displaystyle{ H \subseteq P }[/math]. The first statement implies the second by taking [math]\displaystyle{ H = \lt x\gt }[/math].

GROUPS OF ORDER 15

If [math]\displaystyle{ |G| = 15 }[/math] then [math]\displaystyle{ n_3 \equiv 0\ \mod\ 15 }[/math] and [math]\displaystyle{ n_3 \equiv 1\ \mod\ 3 }[/math]. These imply [math]\displaystyle{ n_3 = 1 }[/math]. Moreover, [math]\displaystyle{ n_5 \equiv 0\ \mod\ 15 }[/math] and [math]\displaystyle{ n_5 \equiv 1\ \mod\ 5 }[/math]. These imply [math]\displaystyle{ n_5 = 1 }[/math]. Thus we have [math]\displaystyle{ P_3 }[/math] a normal [math]\displaystyle{ 3 }[/math]-subgroup. Moreover, we have [math]\displaystyle{ P_5 }[/math] a normal [math]\displaystyle{ 5 }[/math]-subgroup. This tells us a lot about the group.