A Group of Order 256 with Non-Equal Commutator Subgroup and Set of Commutators, Assignments of Algebra

Download A Group of Order 256 with Non-Equal Commutator Subgroup and Set of Commutators and more Assignments Algebra in PDF only on Docsity! MATH 511A, HW 11: A CERTAIN GROUP OF ORDER 256 In this homework, we will investigate an example (due to Carmichael) of a group of order 256 whose set of commutators is not equal to its commutator subgroup. Let G be the subgroup of S16 generated by the following permutations: z0 = (1, 3)(2, 4) z1 = (5, 7)(6, 8) z2 = (9, 11)(10, 12) z3 = (13, 15)(14, 16) y0 = (1, 3)(5, 7)(9, 11) y1 = (1, 2)(3, 4)(13, 15) y2 = (5, 6)(7, 8)(13, 14)(15, 16) y3 = (9, 10)(11, 12) . 1. (a) Verify that z0, . . . , z3 lie in the center of G. (b) Check that z0, . . . , z3 generate a subgroup of order 16, and deduce that the order of Z(G) is at least 16. 2. For any i, j, verify that the commutator [yi, yj ] is an element of {1, z0, . . . , z3}, and that each zk occurs as such a commutator. Specifically, check that [y0, y1] = z0 [y0, y2] = z1 [y0, y3] = z2 [y1, y2] = z3 [y1, y3] = 1 [y2, y3] = 1 . (Be sure to explain why it suffices to check just these six commutators.) Conclude that G′ = 〈z0, z1, z2, z3〉, so #G′ = 16. 3. Show that G/〈z0, z1, z2, z3〉 is a commutative group of order at most 16, and argue that it has order exactly 16 if and only if no element of the form y00 y 1 1 y 2 2 y 3 3 for i ∈ {0, 1} and at least one  6= 0 lies in 〈z0, z1, z2, z3〉 (in which case these elements form a set of coset representatives for 〈z0, z1, z2, z3〉). 4. Consider two elements of the form y00 y 1 1 y 2 2 y 3 3 , y δ0 0 y δ1 1 y δ2 2 y δ3 3 for i, δi ∈ {0, 1}.