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CHAPTER 7

Cosets and Lagrange’ Theorem

Properties of Definition ( of H in ). Let G be a and H G. For all a G, the ah h H is denoted ✓ 2 {1 | 2 }1 by aH. Analagously, = 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 = 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 with addition. Let H be a which is a line through the origin, .., H = tx and = 1 . { | 2 k k }

Then the left coset + H = v + x x H and the right coset { | 2 } H + v = x + v x H are the same line, and is to H. { | 2 } Lemma (Properties of Cosets). Let H G, and let a, 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 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 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 . Then | | | | 2 G a k a k k a| | = a| | = (a| |) = e = e. ⇤ Corollary (5 — Fermat’s Little Theorem.). For all a 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, theresult 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 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 .] ⇡ 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. Now suppose|h bi \=h2.i| Then by|hTheoremi| 7.2 h ai 6 b h=i |ah i \b h=i|p2 > 2p = G , which is| |imp6 ossible. Thus b = 2. |h ih i| |h i||h i| | | | | Now ab a = ab = 2 by the same argument as above. Then 62 h i ) | | 1 1 1 1 ab = (ab) = b a = ba . This relation completely determines the multiplication table for G. [For exam- ple, 3 4 2 4 2 1 4 3 a ba = a (ab)a = a (ba )a = a(ab)a = 1 3 2 1 2 a(ba )a = (ab)a = (ba )a = ba.] Thus all noncyclic groups of order 2p must be isomorphic to each other, and Dp is one such group. ⇤ Example. S D . 3 ⇡ 3 Proof. S = 6 = 2 3 and S is not cyclic. | 3| · 3 ⇤ 7. COSETS AND LAGRANGE’S THEOREM 101 An Application of Cosets to Groups Definition (Stabilizer of a Point). Let G be a group of of a set S. For each i S, let stabG(i) = G (i) = i . We call stabG(i) the stabilizer of i in . { 2 | }

Corollary. stabG(i) is a subgroup of G. Proof. [Page 120 # 35] Let ↵, stab (i). Then ↵(i) = i and (i) = i, so 2 G (↵)(i) = ↵ (i) = ↵(i) = i, so ↵ stab (i). Also, 2 G 1 1 1 ↵ a(i) = ↵ (i) = i = ↵ (i), ) 1 so ↵ stab (i). 2 G Thus stab (i) G by the two-step test. G  ⇤ Definition (Orbit of a Point). Let G be a group of permutations of a set S. For each i S, let orbG(s) = (s) G . The set orbG(s) is a subset of S called the2orbit of s under G.{We use| 2orb} (s) to denote the number of | G | elements in orbG(s). 102 7. COSETS AND LAGRANGE’S THEOREM Problem (Page 158 # 45). Let G = (1), (1 2)(3 4), (1 2 3 4)(5 6), (1 3)(2 4), (1 4 3 2)(5 6), { (5 6)(1 3), (1 4)(2 3), (2 4)(5 6) } Solution. stab (1) = (1), (2 4)(5 6) . G { } stab (3) = (1), (2 4)(5 6) . G { } stab (5) = (1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3) . G { } orb (1) = 1, 2, 3, 4 . G { } orb (3) = 1, 2, 3, 4 . G { } orb (5) = 5, 6 . G { } ⇤ 7. COSETS AND LAGRANGE’S THEOREM 103 Theorem (7.4 — Orbit-Stabilizer Theorem). Let G be a group of permu- tations of a set S. Then, for any i S, G = orbG(i) stabG(i) . Proof. 2 | | | | · | | G By Lagange’s Theorem, | | is the number of distinct left cosets of stab (i) | G | stabG(i) in G. [To show a 1–1 correspondence between these cosets and the elements of orbG(i).] Define T : stab (i) G orb (i) by T ( stab (i)) = (i). { G | 2 } ! G G [To show T is well-defined, i.e., that ↵ stabG(i) = stabG(i) = ↵(i) = (i) or that the result of the mapping depends on the coset and not )on a particular representation of it.] Now, 1 ↵ stab (i) = stab (i) stab (i) = ↵ stab (i) G G () G G () 1 1 ↵ stab (i) ↵ (i) = i (i) = ↵(i) 2 G () () so T is well-defined and, from the reverse implications, T is 1–1. [To show T is onto.] Let orb (i). Then ↵(i) = j for some ↵ G, and 2 G 2 T (↵ stabG(i)) = ↵(i) = j, so T is onto. Thus T is a 1–1 correspondence. ⇤ The Rotation Group of a Cube Let G be the rotation group of a cube. Label the faces 1–6. Since each rotation must carry each face to exactly one other face, G is a group of permutations on 1, 2, 3, 4, 5, 6 .There is a central horizontal or vertical permutation that carries{ face 1 to an} other face, so orb (1) = 6. Also, there are 4 rotations | G | (0, 90, 180, 270 about the line to the face and passing through its ), so stab (1) = 4. By Theorem 7.4,? | G | G = orb (1) stab (1) = 6 4 = 24. | | | G | · | G | · 104 7. COSETS AND LAGRANGE’S THEOREM

Theorem (The rotation Group of the Cube). The group of rotations of a cube is isomorphic to S4. Proof.

Since G = S4 , we need only show that G is isomorphic to a subgroup of S4. Now a| cub| e |has| 4 diagonals and any rotation induces a permutation of these diagonals. But we cannot just assume that di↵erent rotations correspond to di↵erent rotations.

↵ = (1 2 3 4) = (1 4 3 2) We need to show all 24 permutations of the diagonals come from rotations. Numbering the diagonals as above, we see two perpendicular axes where 90 rotations give the permutations ↵ = (1 2 3 4) and = (1 4 3 2). These induce an 8 element subgroup ", ↵, ↵2, ↵3, 2, 2↵, 2↵2, 2↵3 and the 3 element subgroup ", ↵, (↵)2 {. Thus the rotations induce all 24} permutations since { } 24 = lcm(8, 3). ⇤ 7. COSETS AND LAGRANGE’S THEOREM 105 Problem (Page 158 # 46). Prove that a group G of order 12 must have an element of order 2. Proof. Let a G, a = e. By Lagrange’s theorem, a = 12, 6, 4, 3, or 2. 2 6 | | If a = 12, a6 = 2. If a = 6, a3 = 2. If a = 4, a2 = 2. | | | | | | | | | | | | Suppose all non-identity elements of G have order 3. But such elements come in pairs, e.g., if a = 3, a2 = 3. But there are 11 non-identity elements, a | | | | contradiction. Thus G has an element of order 2. ⇤ Problem (Page 158 # 47). Show that in a group G of odd order, the equation x2 = a has a unique solution. Proof. Suppose y is a solution also, i.e., y2 = a = x2 = y2. Let G = 2k + 1. Then ) | | x = xe = xx2k+1 = x2k+2 = (x2)k+1 = (y2)k+1 = y2k+2 = yy2k+1 = ye = y. ⇤