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Math 412. Cosets and the Proof of Lagrange's Theorem: Answers

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Math 412. Cosets and the Proof of Lagrange's Theorem: Answers Math 412. Cosets and the Proof of Lagrange’s Theorem: Answers DEFINITION: Fix a group G and a subgroup K.A right K-coset of K is any subset of G of the form K ◦ b = fk ◦ b j k 2 Kg where b 2 G. Similarly, a left K-coset of K is any set of the form b ◦ K = fb ◦ k j k 2 Kg: PROPOSITION: Fix a group G and a subgroup K. The total number of right K-cosets is equal to the total number of left K-cosets. DEFINITION: Fix a group G and a subgroup K. The index of K in G is the total number of distinct right K-cosets of K in G. We write this index [G : K]. LAGRANGE’S THEOREM: Fix a group G and a subgroup K. Then jGj = jKj[G : K]. A. BASICS OF COSETS. Fix a group G and a subgroup K. Let a; b 2 G. (1) Prove a 2 Ka. Prove also that a 2 aK. Thus the “right K-coset containing a” is the set K ◦ a. Similarly, the “left K coset containing a” is the set a ◦ K. (2) Prove that if a 2 Kb, then also b 2 Ka. (3) Prove that if a 2 Kb and b 2 Kc, then a 2 Kc. (4) Conclude that the relation between two elements a; b 2 G given by “a is in the (right) coset Kb” is an equivalence relation. (5) Show that for all a; b 2 G, either Ka = Kb or Ka \ Kb = ; (but not both).1 (1) We can write a = e ◦ a, and also a = a ◦ e. (2) Say a 2 Kb. So we can write a = k ◦ b for some k 2 K. Hit both sides on the left with k−1 to get ka−1 = b. So b 2 Ka. (3) Say a 2 Kb and b 2 Kc. Write a = k1b and b = k2c. Then a = k1b = k1(k2c) = (k1k2)c; showing thatsa 2 Kc. (4) The items (1), (2), (3) show the relation is reflexive, symmetric, and transitive. So it is an equivalence relation. (5) It suffices to show that if Ka \ Kb is not empty, then Ka = Kb. Say g is in the intersection. Write g = k1a = k2b −1 where k1; k2 2 K. Let us first show that Ka ⊂ Kb. Take arbitrary ka 2 Ka. We know that a = k1 k2b, so −1 substituting, we have ka = kk1 k2b, which shows ka 2 Kb. The reverse inclusion is similar. B. EXAMPLE IN THE GROUP OF INTEGERS Let G = (Z; +) and let K = 7Z be the subgroup of multiples of 7. (1) Make sure you see why 7Z really is a subgroup of Z. What is the operation ◦ in this case? (2) Describe the right K-coset containing 0. (3) Explain why the right K-coset containing a is the same as the set [a]7 ⊂ Z. (4) Find the index [G : K]: Verify LaGrange’s theorem. (1) The group operation is addition. Note that adding two multiples of 7 produces a multiple of 7 (7m+7n = 7(m+n)) and that the inverse of a multiple of 7 is also a multiple of 7 (the inverse of 7n is 7(−n)). (2) The left and right K-coset containing 0 is K = 7Z in both cases. (3) The notation Ka in this context is a + 7Z, which is the set a + 7n j n 2 Zg = [a]7: (4) Each coset of K is an equivalence class mod 7. There are 7 of these. So [G : K] = 7. Lagrange tells us 1 · 7 = 1. C. EXAMPLE IN S3. (1) Write out all six element of S3 in cycle notation. Use e for the identity. (2) Consider the group K ⊂ S3 generated by (1 2). List out all elements of K. What does Lagrange’s theorem predict about the number of right cosets of K? 1You have done a similar exercise multiple times; this is really a general property of any equivalence relation. (3) Find the right K-coset containing e. Find the left coset containing e. (4) Partition S3 up into right K-cosets. Compute [S3 : K]. Verify Lagrange’s Theorem in this case. (5) Partition S3 up into left K-cosets. Verify that we have the same number of left and right cosets. (1) S3 = fe; (12); (23); (13); (123); (132)g. (2) K = fe; (12)g: Since jKj = 2 and jS3j = 6, Lagrange predicts [S3 : K] = 3, so three right cosets. (3) In both cases Ke = eK = K = fe; (12)g: (4) The three right K-cosets are • Ke = K(12) = fe; (12)g, • K(13) = K(132) = f(13); (132)g and • K(23) = K(123) = f(23); (123)g. So there are 3 cosets, meaning [S3 : K] = 3. Since jKj = 2 and S3j = 6, Lagrange’s Theorem is verified. (5) The three left K-cosets are eK = (12)K = K = fe; (12)g, (13)K = (132)K = f(13); (132)g and (23)K = (123)K = f(23); (123)g. D. EXAMPLE IN THE SYMMETRY GROUP OF THE SQUARE. Let G be the dihedral group D4. Let R4 be the subgroup generated by r, the rotation 900 counterclockwise. (1) List out all elements in the right coset R4r containing r. (2) List out all elements in the coset R4a containing a. (3) Describe the partition of D4 up into right R4-cosets. Is there a geometric interpretation or description of each? (4) Verify Lagrange’s Theorem. 2 3 (1) R4r = fr; r ; r ; eg. (2) R4a = fa; d; x; yg. (3) There are two right cosets, one is the set of all rotations R4e, including the null-rotation. The others is the set of all reflections. Both have four elements. (4) jD4j = 8 and jR4j = 4, so [D4 : R4] = 2. E. THE PROOF OF LAGRANGE’S THEOREM. Fix a group G and a subgroup K. Let a; b 2 G. (1) Prove that there is a bijection Ka ! Kb given by right multiplication by a−1b. (2) Prove that all right K-cosets have the same cardinality. (3) Prove that if G is finite, then jGj = [G : K]jKj. (4) Conclude that the order of any subgroup K must divide the order of G. (1) The map Ka ! Kb sending ka to ka ◦ a−1b is a bijection, because it has inverse given by right multiplication by b−1a. (2) (1) shows that all right K-cosets have the same number of elements. (3) We can think of the K-cosets as partitioning up G into [G : K] equal-sized sets of cardinality K. So jGj = jKj[G : K]. (4) Clear. F. ARE LEFT AND RIGHT COSETS THE SAME? Let G be a group and K a subgroup. (1) With notation as in Problem B, is 1 + K = K + 1? What about a + K = K + a for arbitrary a? (2) With notation as in Problem C, is K ◦ (123) = (123) ◦ K? (3) With notation as in Problem D, is R4r = rR4? What about R4a = aR4? (4) TRUE OR FALSE: In an arbitrary group G, with subgroup K, we have K ◦ g = g ◦ K for all g. (5) TRUE OR FALSE: In an arbitrary group G, we have K ◦ eG = eG ◦ K for all subgroups K. (6) TRUE OR FALSE: In an arbitrary abelian group G, with subgroup K, we have K ◦ g = g ◦ K for all g. (7) TRUE OR FALSE: In an arbitrary group G, every right K-coset is a subgroup of G. (8) TRUE OR FALSE: In an arbitrary group G, the right K-coset containing the identity element is a subgroup of G. (1) Yes, because Z is abelian a + K = K + a for any subgroup. (2) No! K ◦ (123) = f(123); (23)g but (123) ◦ K = f(123); (13)g: 2 3 (3) Yes! R4r = rR4 = fe; r; r ; r g. Since Ra and aR also both have four elements, and are disjoint from Rr, they both have no choice but to be the full set of 4 reflections. (4) False! See 2. (5) True! e commutes with everything. (6) True! (7) False! e belongs to only one of the cosets! A subgroup must contain e. (8) True. Ke = K. 1 17 G. EXAMPLE IN MATRIX GROUPS. Fix G = GL ( ) and the subgroup K = SL ( ). Let A = . 2 R 2 R 0 π (1) Prove that the right K-coset KA in GL2(R) is fB 2 GL2(R) j det B = πg. (2) Prove that the left K-coset AK = KA. (3) Prove that the right K-cosets KC and KD are the same in this case if and only if det C = det D. (4) What is the index [GL2(R): SL2(R)]? −1 −1 −1 −1 (1) Say B 2 KA. So BA 2 SL2. This means det(BA ) = 1. So det(B) det(A ) = det(B)(det(A)) = 1, or det B = det A = π. So KA ⊂ fB 2 GL2(R) j det B = πg. For the other direction, take B with determinant π. −1 −1 −1 −1 We can write B = BA A, and observe that det BA = det(det A) = 1, so BA 2 SL2 and B 2 KA. (2) Note that both AA−1 and A−1A have deteminant one and are inverse to eachother. Take arbitrary element in KA. It 0 0 is of the form BA for some matrix B with determinant one. We need to write it in the form AB for some B 2 SL2.
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