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1. Group Action; Definition and First Theorems: 1.1. Definition. We Start by Recalling the Notion of an Equivalence Relation. Le

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1. Group Action; Definition and First Theorems: 1.1. Definition. We Start by Recalling the Notion of an Equivalence Relation. Le 1. Group action; Definition and first theorems: 1.1. Definition. We start by recalling the notion of an equivalence relation. Let A be a set. A relation R on A is a subset of A A.If(a, b) R then we write aRb and we say that a is related to b via the relation R. We× simply say a is∈ related to b if the relation R is understood from context. Let A be a set and R be a relation on A.WesaythatR is an equivalence relation on A if R satisfies the following three conditions: (i) aRa for all a A, (ii) If aRb,thenbRa for all a, b A, (iii) If aRb and bRc,thenaRc for all a, b, c A. ∈ Let R∈ be an equivalence relation on a set A.Ifa A∈,wedefinetheequivalence class of a to be the set of all b A such that a is related to b∈. Verify that: ∈ If two elements a, b A are related, then their equivalence classes are identical, ◦ otherwise their equivalence∈ classe are disjoint. members of the same equivalence class are all related to each other and members of ◦ distinct equivalence classes are never related to each other. The set A is the disjoint union of all the equivalence classes. ◦ 1.2. Definition. Let G be a group with identity element e and A be a set. A left action of G on A is a map ρ : G A A such that ρ(e, a)=a for all a A and ρ(g,ρ(h, a)) = ρ(gh,a) for all g,h G and×a →A. Usually we shall use the notation∈ ρ(g,a)=g.a.Sowehave e.a = a for all∈ a A and∈ g.(h.a)=(gh).a, for all a A and for all g,h G. ∈ ∈ ∈ 1.3.Exercise. Show that specifying an action of G on A is equivalent to having a group homomorphism from G to Aut(A), the group of all one to one and onto maps from A to A. 1.4.Examples: (1) Sn acts on 1, 2, ,n . { ···n } (2) GLn(R)actsonR . (3) More generally if G is a group of bijections of a set A,thenG acts on A. (4) a group G acts on itself by left translation, by conjugation. (5) a group acts G on the set of left cosets of a subgroup H by left translation.. (6) a group acts on the set of subsets of a particular order by left translation. (7) a group G acts on the set of subgroups of G, on the set of subgroups of a particular order, by conjugation. (8) Rotational symmetries of the cube acts on the cube, acts on the set of vertices of the cube, the set of diagonals of the cube... (9) Symmetries of the icosahedron acts on 1, 2, 3, 4, 5 . { } (10) SL2(Z) acts on the upper half plane. 1.5. Definition. Suppose G acts on A.Ifa, b A, we write a b if there exists g G ∈ ∼G ∈ such that ga = b. Verify that G is an equivalence relation on A. The equivalence class of a A is called the orbit of a under∼ G and is denoted by .So ∈ Oa = b A: b = ga for some g G . Oa { ∈ ∈ } We write (G A) for the set of orbits. Since G is an equivalence relation, A is the disjoint union of the\ orbits, that is, ∼ A = , O O∈!G\A 1 where denotes disjoint union. This is called the orbit decomposition of A.Fora A, let ∈ " G = g G: ga = a . a { ∈ } Verify that Ga is a subgroup of G.ThesubgroupGa is called the stabilizer of a in G. 1.6. Theorem (Lagrange). Let H be a subgroup of a group G.ConsiderH acting on G by left multiplication: H G G given by (h, g) hg.Theorbitofg G is the set Hg = hg : h H .Theseorbitsarecalledtherightcosetsof× → &→ H in G.Thesetofright∈ cosets of{ H in∈G is} denoted by H G.ThegroupG is the disjoint union of the right cosets of H.IfG is finite, then one has \G = H H G . | | | || \ | Proof. Consider H acting on G by left multiplication. From orbit decompostion, we get that G is a disjoint union of the right cosets: G = . O O∈!H\G Verify that each coset of H in G has the same cardinality, namely H .IfG is finite, it follows that G = H H G . | | ! | | | || \ | 1.7.Definition/Remark: Let G be a group and A be a set. A right action of G on A is amapA G A written (a, g) a.g such that a.e = a and (a.g).h = a.(gh) for all a A and g,h ×G.If→ G acts on A on&→ the right, then the set of orbits will be denoted by A/G∈ . Everything∈ we said about left action carries over verbatim for right actions. If H is a subgroup of G, then there is also a right action of H on G via right multiplication: G H G given by (g,h) gh. The orbits have the form gH for g G and are called left× cosets→ of H in G. So the&→ set of left cosets of H in G is denoted by G/H∈ . The obvious analog of theorem 1.6 holds for the set of left cosets. When we say G acts on A, we shall mean G acts on the left, unless otherwise specified. 1.8. Theorem (Orbit-Stabilizer theorem). Let G be a group acting on a set A.Fixa A. ∈ Then one has a bijection G/Ga a.InparticularG = Ga a .SoG is finite if and only if G and are finite for some≃ O a A. | | | ||O | a Oa ∈ Proof. Consider the map ϕ : G given by ϕ(g)=ga. By definition of orbit, this is an → Oa onto map from G to a. Claim: If g G,thenOϕ−1(ga)=gG . The verification is routine: ∈ a h ϕ−1(ga) ϕ(h)=ga ha = ga g−1ha = a g−1h G ∈ ⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒ ∈ a h gG . ⇐⇒ ∈ a Note that each element b a have the form b = ga for some g G,soforeachsuchb a, −1 ∈ O ∈ ∈ O the preimage ϕ (b) is a left coset of Ga in G.ThuswehavethemapΦ : a G/Ga given by Φ(b)=ϕ−1(b). So for all g G,wehaveΦ(ga)=gG . Verify that ΦO: → G/G is ∈ a Oa → a one to one and onto. In particular G/Ga = a . Now Lagranges theorem (i.e. the analog of theorem 1.6 for left cosets) implies| that| G|O | = G G/G = G . ! | a||Oa| | a|| a| | | −1 1.9. Example. Let G be a group and g,h G.Definecg(h)=ghg . The operation g −1 ∈ &→ ghg is called conjugating h by g. Verify that the map G G G given by (g,h) cg(h) is a left action of G on G, this is called the conjugation action.× → Consider G acting on&→ G by conjugation. The orbit of an element a G under this action consists of all the elements of ∈ 2 the form gag−1,whereg varies over G. This orbit is called the conjugacy class of a,andwe shall denote it by Class(a). The orbit decomposition implies that the G is the disjoint union of all the conjugacy classes. Let a G. The stabilizer of G under the conjugation action is g G: gag−1 = a . ∈ { ∈ } This subgroup is called the centralizer of a in G and is denoted by CG(a). Now assume that G is finite. Then the orbit stabilizer theorem implies that for each a G,wehave ∈ G = CG(a) Class(a) . | Recall| | that|| the center| of G,denotedZ(G), consists of all the elements z G such that ∈ zg = gz for all g G.LetZ(G)= z1, ,zm .Inotherwords, z1 , , zm are precisely the one element∈ conjugacy classes{ in G···.Let } , , be the rest{ } of··· the{ conjugacy} classes O1 ··· Or and choose a1 1, ,ar r,so j = Class(aj). We say that aj is a representative for the conjugacy∈ classO ···. Now∈ fromO orbitO decomposition, we get that G is a disoint union of Oj the sets z1 , , zm , Class(a1), , Class(ar). So { } ··· { } ··· r G = Z(G) + Class(a ) . | | | | | j | #j=1 Using the orbit stabilizer theorem we get, the class equation: r G G = Z(G) + | | . | | | | C (a ) #j=1 | G j | 3 .
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