Part 3. 1 Part 3. 2 A Little Group Theory • The element b in (3) is unique: Suppose • A Group is a set G equipped with a ab = ba = e binary operation G × G:(a, b) 7→ ab such ab0 = b0a = e that the following properties hold. Then 1. a(bc) = a(bc). (Associative Law) b0ab = (b0a)b = eb = b 2. The is an e ∈ G such that ea = ae = a b0ab = b0(ab) = b0e = b0 for all a ∈ G. (Existence of identity) 0 3. For every a ∈ G there is a b ∈ G such so b = b. This unique element is −1 that ab = T BA = e. (Existence of denoted b . inverses) • A subset H ⊆ G is a subgroup of G if it is a group under the binary operation of • There is only one identity element: G, i.e., Suppose ea = ae = a and e0a = ae0 = a for all a ∈ G. Then e = ee0 = e0. 1. e ∈ H 2. a, b ∈ H =⇒ ab ∈ H. 3. a ∈ H =⇒ a−1 ∈ H • If H1 and H2 are subgroups of G, so is H1 ∩ H2. • We write H ≤ G to indicate that H is a subgroup of G. Part 3. 3 Part 3. 4 • G ⊆ G and { e } ⊆ G are subgroups. • If a ∈ G then hai is a subgroup. It’s clear that • If S ⊆ G, the set { H | H ≤ G, H ⊇ S } is non-empty because it contains G. Set hai = { an | n = 0, ±1, ±2,... }. \ (By definition a0 = e and a−k for k > 0 hSi = { H | H ≤ G, H ⊇ S } means (a−1)k.) Then hSi is a subgroup of G. It is the There are two possibilities: Either all of smallest subgroup of G that contains S, the powers an are distinct, or two of i.e., S ⊆ hGi and if H ≤ G and H ⊇ S, them are the same. Suppose am = an then hSi ⊆ H. The subgroup hSi is and choose the notation so that m > n. called the subgroup of G generated by Then m = n + k where k > 0. Then S. If hSi = G, we say that S generates an = am = an+k = anak, so an = anak. G. Multiplying this equation on the left by a−n gives ak = e. Thus, some power of a is the identity. Let p be the smallest positive integer so that ap = e. We call p the order of a, denoted by o(a). In this case, hai is finite, namely hai = { e, a, a2, . , ap−1 }. • A group G is cyclic if G = hai for some a ∈ G. We say a is a generator of G. Part 3. 5 Part 3. 6 Equivalence Relations Suppose x ∼ y. Let z be an element of [x]. Then z ∼ x; combining this with • Let X be a nonempty set. A relation ∼ x ∼ y we get z ∼ y. Thus, z ∈ [y]. This on X is an equivalence relation if it shows [x] ⊆ [y]. Similarly, [y] ⊆ [x], so satisfies the following properties. [x] = [y]. 1. x ∼ x for all x ∈ X. (Reflexive) Suppose that [x] ∩ [y] 6= ∅. Then there is 2. x ∼ y =⇒ y ∼ x. (Symmetric) some z ∈ [x] ∩ [y]. But this means that 3. x ∼ y, y ∼ z =⇒ x ∼ z. (Transitive) z ∼ x and z ∼ y. But then x ∼ y, so [x] = [y]. • If x ∈ X we define [x], the equivalence class of x by Each x ∈ X is in some equivalence class (namely [x]) and the equivalence classes [x] = { y | y ∼ x }. are disjoint, so we have X described as a union of a collection of disjoint Since x ∼ x, x ∈ [x]. subsets. That’s what it means to • Proposition partition X. 1. [x] = [y] if and only if x ∼ y. 2. Either [x] = [y] or [x] ∩ [y] = ∅. 3. The equivalence classes partition X. • Proof: If [x] = [y] then x ∈ [x] = [y], so x ∈ [y]. By the definition of [y], x ∼ y. Part 3. 7 Part 3. 8 Cosets This is called the left coset of a modulo H. Thus, G is the disjoint union of the • Let H be a subgroup of G. Define a left cosets. The collection of left cosets relation ∼ on G by a ∼ b if there is some modulo H is called G/H. h ∈ H so that ah = b. We claim this is an equivalence relation. • We can similarly define a relation ∼ by a ∼ b if there is an element h of H so If a ∈ G then ae = a and e ∈ H so a ∼ a. that ha = b. The equivalence class of a Suppose that a ∼ b. Then there is some with respect to this relation is [a] = Ha, h ∈ H so that ah = b. Multiplying this which is called the right coset of a equation on the right by h−1 gives modulo H. The collection of right bh−1 = ahh−1 = ae = a. Since h−1 ∈ H, cosets is called H \ G. b ∼ a. Suppose that a ∼ b and b ∼ c. Then • A group is called finite if it has only finitely many elements. there are elements h1, h2 ∈ H such that ah1 = b and bh2 = c. Multiply the • |X| denotes the number of elements in equation ah1 = b on the right by h2. X. If G is a group, |G| is often called This gives ah1h2 = bh2 = c. Thus, the order of G. ah1h2 = c. Since h1h2 ∈ H, we get a ∼ c. • What is [a]? [a] = { ah | h ∈ H } = aH. Part 3. 9 Part 3. 10 • Let G be a group and H a finite • It’s possible that G/H is finite even if G subgroup. We can define a 1-1 and and H are infinite. The number of onto map f : H → aH by f(h) = ah. elements in G/H is often denoted Thus, H and aH are in 1-1 [G : H], called the index of H in G. correspondence, so |aH| = |H|, i.e., An Example every left coset has the same number of elements as H. Similarly, every right • Let Z = { 0, ±1, ±2, ±3 ... } be the set of coset has the same number of elements integers. This is a group under the as H. operation of addition. In this case the group is commutative. • Lagrange’s Theorem Let G be a finite group and let H be a subgroup. Then • Let n be a positive integer and write nZ = { nk | k ∈ Z } = { 0, ±n, ±2n, ±3n, . }, |G/H| |H| = |G|. i.e., nZ is the set of all multiples of n. It In particular, |H| divides |G| and should be easy to see that nZ is a subgroup of . |G| Z |G/H| = . |H| • Since Z is commutative, there’s really no difference between right and left Similarly, cosets. The relation for the cosets is |G| a ∼ b if there is an h ∈ n so that |H \ G| = . Z |H| a + h = b. In other words b − a = nk for Part 3. 11 Part 3. 12 some k ∈ Z. Another way to say it then this: If [r0] = [r] and [s0] = [s], is it true is that a ∼ b if b − a is divisible by n. that [r0 + s0] = [r + s]? If not, we would The equivalence class of a is get a different answer for the sum of two cosets depending on which [a] = a + nZ = { a + nk | k ∈ Z }. elements of the cosets we choose to The set of equivalence classes is represent them. denoted by Z/nZ (read “ Z mod n Z”) Fortunately, the required property holds. or Zn (read “Z mod n”). The distance If [r0] = [r] then r0 ∼ r, equivalently, elements of can be listed as 0 0 Zn r ∼ r , so r = r + nk for some k ∈ Z. 0 0 [0], [1], [2],..., [n − 1], Similarly, if [s ] = [s], then s = s + n` for some ` ∈ Z. But then for k ∈ Z, [k] must be one of the elements of the above list. (How do you r0 + s0 = r + nk + s + n` = (r + s) + n(k + `). determine which one?) Since k + ` ∈ Z, this shows that • We show that Zn can be made into a (r0 + s0) ∼ (r + s) so [r0 + s0] = [r + s]. group by defining the group operation Now that the operation makes sense, by the group properties follow easily form [r] + [s] = [r + s], r, s ∈ Z. the group properties of Z. The main point is to show that this definition makes sense! The problem is Part 3. 13 Part 3. 14 For example, for a, b, c ∈ Z, Normal Subgroups [a] + ([b] + [c]) = [a] + [b + c] • In the case of a noncommutative group, an additional condition is required to = [a + (b + c)] make G/H a group. = [(a + b) + c] • Let G be a group and H an subgroup. = [a + b] + [c] If g ∈ G, we define = ([a] + [b]) + [c], g−1Hg = { g−1hg | h ∈ H }. where we have used the associative law for Z. Thus Zn is associative. • H is called a normal subgroup of G if We have [0] + [a] = [0 + a] = [a], so [0] is −1 the identity element. g Hg ⊆ H, for all g ∈ G. We then have [a] + [−a] = [a + (−a)] = [0], We write H E G to indicate H is a so [−a] is the inverse of [a]. normal subgroup of G. Part 3. 15 Part 3. 16 • If H E G, then gH = Hg for all g ∈ G, • If H is a normal subgroup of G, the i.e., there’s no difference between the collection of cosets G/H can be made left coset and the right coset.