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

DEFINITION: Fix a G and a K.A right K-coset of K is any subset of G of the form K ◦ b = {k ◦ b | k ∈ K} where b ∈ G. Similarly, a left K-coset of K is any set of the form b ◦ K = {b ◦ k | k ∈ K}.

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 |G| = |K|[G : K].

A.BASICSOF COSETS. Fix a group G and a subgroup K. Let a, b ∈ G. (1) Prove a ∈ Ka. Prove also that a ∈ 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 ∈ Kb, then also b ∈ Ka. (3) Prove that if a ∈ Kb and b ∈ Kc, then a ∈ Kc. (4) Conclude that the relation between two elements a, b ∈ G given by “a is in the (right) coset Kb” is an . (5) Show that for all a, b ∈ 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 ∈ Kb. So we can write a = k ◦ b for some k ∈ K. Hit both sides on the left with k−1 to get ka−1 = b. So b ∈ Ka. (3) Say a ∈ Kb and b ∈ Kc. Write a = k1b and b = k2c. Then a = k1b = k1(k2c) = (k1k2)c, showing thatsa ∈ 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 ∈ K. Let us first show that Ka ⊂ Kb. Take arbitrary ka ∈ Ka. We know that a = k1 k2b, so −1 substituting, we have ka = kk1 k2b, which shows ka ∈ Kb. The reverse inclusion is similar.

B.EXAMPLEINTHE GROUPOF 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 | n ∈ Z} = [a]7. (4) Each coset of K is an mod 7. There are 7 of these. So [G : K] = 7. Lagrange tells us ∞ · 7 = ∞.

C.EXAMPLEIN 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 = {e, (12), (23), (13), (123), (132)}. (2) K = {e, (12)}. Since |K| = 2 and |S3| = 6, Lagrange predicts [S3 : K] = 3, so three right cosets. (3) In both cases Ke = eK = K = {e, (12)}. (4) The three right K-cosets are • Ke = K(12) = {e, (12)}, • K(13) = K(132) = {(13), (132)} and • K(23) = K(123) = {(23), (123)}. So there are 3 cosets, meaning [S3 : K] = 3. Since |K| = 2 and S3| = 6, Lagrange’s Theorem is verified. (5) The three left K-cosets are eK = (12)K = K = {e, (12)}, (13)K = (132)K = {(13), (132)} and (23)K = (123)K = {(23), (123)}.

D.EXAMPLEINTHE SYMMETRY GROUPOFTHE 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 = {r, r , r , e}. (2) R4a = {a, d, x, y}. (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) |D4| = 8 and |R4| = 4, so [D4 : R4] = 2.

E.THEPROOFOF LAGRANGE’S THEOREM. Fix a group G and a subgroup K. Let a, b ∈ 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 . (3) Prove that if G is finite, then |G| = [G : K]|K|. (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 |G| = |K|[G : K]. (4) Clear.

F. ARE LEFTAND RIGHT COSETSTHE 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) TRUEOR FALSE: In an arbitrary group G, with subgroup K, we have K ◦ g = g ◦ K for all g. (5) TRUEOR FALSE: In an arbitrary group G, we have K ◦ eG = eG ◦ K for all K. (6) TRUEOR FALSE: In an arbitrary G, with subgroup K, we have K ◦ g = g ◦ K for all g. (7) TRUEOR FALSE: In an arbitrary group G, every right K-coset is a subgroup of G. (8) TRUEOR 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) = {(123), (23)} but (123) ◦ K = {(123), (13)}. 2 3 (3) Yes! R4r = rR4 = {e, r, r , r }. 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.EXAMPLEIN 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 {B ∈ GL2(R) | det B = π}. (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 ∈ KA. So BA ∈ SL2. This means det(BA ) = 1. So det(B) det(A ) = det(B)(det(A)) = 1, or det B = det A = π. So KA ⊂ {B ∈ GL2(R) | det B = π}. 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 ∈ SL2 and B ∈ 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 ∈ SL2. Multiply BA on the left by AA−1 and on the right by A−1A, which of course leaves the element the same since this is multiplying on the left/right by the identity. So BA = AA−1BAA−1A = A(A−1BA). This will be in AK −1 if A BA is in K = SL2, which it is, since determinant is multiplicative. −1 −1 (3) KC = KD if and only if C D ∈ SL2. So if and only if det C(det D) = 1, or det C = det D. (4) ∞.

G.CAUTIONARY EXAMPLE. Let G be a group and let K be a subgroup. Consider the set G/K of right K-cosets. It is tempting to try to define a group as we defined quotient rings. That is, we can try to define a binary operation ? on G/K by (K ◦ g) ? (K ◦ h) = K ◦ (g ◦ h). (1) Show that using the example of Z 7 in Z from (B), that ? gives a well defined binary operation, which satisfies the axioms of a group. What is this group? (2) Show that using the example of K = h(1 2)i in S3 from (C), that ? does not gives a well defined binary operation. Thus there is no way to induce a structure on the set of cosets G/K.! (3) Show that using the example of R4 in D4 from (D), that ? does give a well defined binary operation. Describe it!

(1) This is exactly the well-definedness for the addition in the Z/7Z = Z7. (2) Let’s try to multiple the coset K(13) = {(13), (123)} times itself. But notice: (13)(13) = e, where as (123)(123) = (132). So is the answer Ke = {e, (12)} or K(132) = {(132), (13)}? We do not always land in the same coset! (3) We have seen that any rotation composed with any rotation is a rotation, and that any reflection composed with any reflection is a rotation, that any reflection composed with any rotation is a reflection, and also that rotation composed with any reflection is a reflection. So no matter which representative of a coset we use, the composition rule on cosets is well-defined.