CHAPTER 7 Cosets and Lagrange’s Theorem Properties of Cosets Definition (Coset of H in G). Let G be a group and H G. For all a G, the set ah h H is denoted ✓ 2 {1 | 2 }1 by aH. Analagously, Ha = ha h H and aHa− = aha− h H . When H G, aH is called {the|left2coset} of H in G containing{ a|, and2 H}a is called the right coset of H in G containing a. In this case the element a is called the coset representative of aH or Ha. aH and Ha are used to denote the number of elements in aH and Ha, respectiv| ely| . | | 91 92 7. COSETS AND LAGRANGE’S THEOREM Example. Consider the left cosets of H = (1), (1 2)(34), (1 3)(2 4), (1 4)(2 3) = ↵ , ↵ , ↵ , ↵ A { } { 1 2 3 4} 4 from the table below: H = 1H = ↵ H = ↵ , ↵ , ↵ , ↵ = ↵ H = ↵ H = ↵ H 1 { 1 2 3 4} 2 3 4 ↵ H = ↵ , ↵ , ↵ , ↵ = ↵ H = ↵ H = ↵ H. 5 { 5 6 7 8} 6 7 8 ↵ H = ↵ , ↵ , ↵ , ↵ = ↵ H, ↵ H, ↵ H. 9 { 9 10 11 12} 10 11 12 Also, for k = 1, 2, . , 12, ↵kH = H↵k. 7. COSETS AND LAGRANGE’S THEOREM 93 When the group operation is addition, we use a + H and H + a instead of aH and Ha. Example. Let G be the group of vectors in the plane with addition. Let H be a subgroup which is a line through the origin, i.e., H = tx t R and x = 1 . { | 2 k k } Then the left coset v + H = v + x x H and the right coset { | 2 } H + v = x + v x H are the same line, and is parallel to H. { | 2 } Lemma (Properties of Cosets). Let H G, and let a, b G. Then 2 (1) a aH. 2 Proof. a = ae aH. 2 ⇤ (2) aH = H a H. Proof. () 2 (= ) Suppose aH = H. Then a = ae aH = H. ) 2 94 7. COSETS AND LAGRANGE’S THEOREM ( =) Now assume a H. Since H is closed, aH H. Next assume h H ( 1 2 ✓ 2 also, so a− h H since H G. Then 2 1 1 h = eh = (aa− )h = a(a− h) aH, 2 so H aH. By mutual inclusion, aH = H. ✓ ⇤ (3) (ab)H = a(bH) and H(ab) = (Ha)b. Proof. Follows from the associative property of group multiplication. ⇤ (4) aH = bH a bH. () 2 Proof. (= ) aH = bH = a = ae aH = bH. ) ) 2 ( =)a bH = a = bh where h H = aH = (bh)H = b(hH) = bH ( 2 ) 2 ) ⇤ (5) aH = bH or aH bH = . \ ; Proof. Suppose aH bH = . Then x aH bH = h , h H \ 6 ; 9 2 \ ) 9 1 2 2 −3− x = ah1 and x = bh2. Thus 1 1 1 1 a = xh1− = bh2h1− and aH = bh2h1− H = b(h2h1− H) = bH by (2). ⇤ 1 (6) aH = bH a− b H () 2 Proof. 1 (2) 1 aH = bH H = a− bH a− b H. () () 2 ⇤ 7. COSETS AND LAGRANGE’S THEOREM 95 (7) aH = bH . | | | | Proof. [Find a map ↵ : aH bH that is 1–1 and onto] ! Consider ↵ : aH bH defined by ↵(ah) = bh. This is clearly onto bH. Suppose ↵(ah ) =!↵(ah ). Then bh = bh = h = h by left cancellation 1 2 1 2 ) 1 2 = ah1 = ah2, so ↵ is 1–1. Since ↵ provides a 1-0-1 correspondence between aH) and bH, aH = bH . | | | | ⇤ 1 (8) aH = Ha H = aHa− . () Proof. 1 1 1 1 aH = Ha (aH)a− = (Ha)a− = H(aa− ) = H aHa− = () () H. ⇤ (9) aH G a H. () 2 Proof. (5) (2) (= ) If aH G, e ah = aH eH = = aH = eH = H = a H. ) 2 ) \ 6 ; ) ) 2 ( =) If a H, aH = H by (2), so aH G since H G. ( 2 ⇤ Note. Analagous results hold for right cosets. Note. From(1), (5), and (7), the left (right) cosets of H partition G into equivalence classes under the relation a b aH = bH ( or Ha = Hb). ⇠ () 96 7. COSETS AND LAGRANGE’S THEOREM Lagrange’s Theorem and Consequences Theorem (7.1 — Lagrange’s Theorem: H Divides G ). If G is a finite group and H G, then H G . Moreover,| | the numb| |er of distinct left | G| | | (right) cosets of H in G is | |. H Proof. | | Let a1H, a2H, . , arH denote the distinct left cosets of H in G. Then, for all a G, aH = aiH for some i = 1, 2, . , r. By (1) of the Lemma, a aH. Th2us 2 G = a H a H a H. 1 [ 2 [ · · · [ r By (5) of the Lemma, this union is disjoint, so G = a H + a H + + a H = r H | | | 1 | | 2 | · · · | r | | | since a H = aH for i = 1, 2, . r. | i | | | ⇤ Definition. The index of a subgroup H in G is the number of distinct left cosets of H in G, and is denoted by G : H . | | G Corollary (1 — G : H = | |.). If G is a finite group and H G, | | H G | | then G : H = | |. | | H | | Proof. Immediate consequence of Lagrange’s Theorem. ⇤ Corollary (2— a Divides G ). In a finite group, the order of each element divides the or|der| of the |group.| Proof. For a G, a = a , so a G . 2 | | |h i| | | | | ⇤ 7. COSETS AND LAGRANGE’S THEOREM 97 Corollary (3 — Groups of Prime Order are Cyclic). A group of prime order is cyclic. Proof. Suppose G has prime order. Let a G, a = e. The a G and a = 1, 2 6 |h i| | | |h i| 6 so a = G = a = G. ⇤ |h i| | | ) h i G Corollary (4 — a| | = e.). Let G be a finite group and a G. Then G 2 a| | = e. Proof. By Corollary 2, G = a k for some k N. Then | | | | 2 G a k a k k a| | = a| | = (a| |) = e = e. ⇤ Corollary (5 — Fermat’s Little Theorem.). For all a Z and every prime p, 2 ap mod p = a mod p. Proof. By the division algorithm, a = pm + r where 0 r < p. Thus a mod p = r, so we need only show rp mod p = r. if r = 0, theresult is clear, so r U(p) = 1, 2, . , p 1 . 2 { − } p 1 p Then, by Corollary 4, r − mod p = 1 = r mod p = r. ) ⇤ 98 7. COSETS AND LAGRANGE’S THEOREM Example. In A4, there are 8 elements of order 3 (↵5–↵12). Suppose H A4 with H = 6. | | 2 Let a A4 with a = 3. Since A4 : H = 2, at most two of H, aH, and a H are distinct.2 If a | |H, H = aH|= a2H,| a contradiction. 2 If a H, then A = H aH = a2 H or a2 aH. 62 4 [ ) 2 2 Now a2 H = (a2)2 = a4 = a H, again a contradiction. 2 ) 2 If a2 aH, a2 = ah for some h H = a H. Thus all 8 elemts of order 3 are in2H, a group of order 6, a con2 tradiction.) 2 Note. This example means the converse of Lagrange’s Theorem is not true. 7. COSETS AND LAGRANGE’S THEOREM 99 H K Theorem (7.2 — HK = | || | ). For two finite subgroups H and | | H K K of a group G, define the set| H\ K | = hk h H, k K . Then H K { | 2 2 } HK = | || | . | | H K | \ | Proof. Although the set HK has H K products, all not need be distinct. That is, | || | we may have hk = h0k0 where h = h0 and k = k0. To determine HK , we need to determine the extent to whic6 h this happ6 ens. For every t | H | K, 1 2 \ hk = (ht)(t− k), so each group element in HK is represented by at least 1 1 H K products in HK. But hk = h0k0 = t = h− h0 = kk0 H K, | \ | 1 ) 2 \ so h0 = ht and k0 = t− k. Thus each element in HK is represented by exactly H K H K products, and so HK = | || | . | \ | | | H K ⇤ | \ | Problem (Page 158 # 41). Let G be a group of order 100 that has a subgroup H of order 25. Prove that every element of order 5 in G is in H. Solution. Let a G and a = 5. Then by Theorem 7.2 the set a H has exactly 5 H2 | | h i · | | elements and a H divides a = 5 = a H = 5 = a H |h i \ | |h i| ) |h i \ | ) |h i \ | a H = a = a H. h i \ h i ) 2 ⇤ 100 7. COSETS AND LAGRANGE’S THEOREM Theorem (7.3 — Classification of Groups of Order 2p.). Let G be a group of order 2p where p is a prime greater than 2. Then G Z2p or G Dp. ⇡ ⇡ Proof. Assume G has no element of order 2p. [To show G D .] ⇡ p [Show G has an element of order p.] By Lagrange’s Theorem, every nonidentity element must have order 2 or p. So assume every nonidentity element has order 2. Then, for all a, b G, 2 1 1 1 (ab) = (ab)− = b− a− = ba, so G is Abelian. Thus, for a, b G, a = e, b = e, e, a, b, ab is closed and so is a subgroup of G of order 4, con2 tradicting6 Lagrange’s6 { Theorem.} Thus G has an element of order p. Call it a. [To show any element not in a has order 2.] Suppose b G, b a . By Lagrange’s theorem and our assumptionh i that G does not ha2ve an 62elemenh i t of order of 2p, b = 2 or b = p. Because a b divides a = p and a = b , |a| b =| |1.