Math 3124 Wednesday, October 15
Fifth Homework Solutions
1. (1 2 3 4 5)(2 3 4 5 6) = (1 2 4)(3 5 6). Therefore the order is [3,3] = 3.
2. Problem 14.33 on page 81. Prove or give a counterexample. If a group has a subgroup
of order n, then it has an element of order n.
S3is a subgroup of S3of order 6. But the elements of S3have order 1, 2 or 3, so we
have a counterexample.
3. Problem 15.2 on page 84. Prove that h24,โ36,54i=h6i.
24 โ h6ibecause 24 =4โ6, โ36 โ h6ibecause โ36 =โ6โ6, and 54 โ h6ibecause
54 =9โ6. Therefore h24,โ36,54iโh6i.
Conversely 6 =24 +2โ(โ36) + 54, so 6 โ h24,โ36,54iand we deduce that h6i โ
h24,โ36,54i. The required equality follows.
4. Problem 15.21 on page 84.
(a) List the elements of S3รZ2.
(b) List the elements of the cyclic subgroup h((1 2),[1])iof S3รZ2.
(c) List the elements of the cyclic subgroup h((123),[1])iof S3รZ2.
(a) ((1), [0]) ((2 3), [0]) ((3 1), [0]) ((1 2), [0]) ((1 2 3), [0]) ((1 3 2), [0])
((1), [1]) ((2 3), [1]) ((3 1), [1]) ((1 2), [1]) ((1 2 3), [1]) ((1 3 2), [1])
(b) ((1), [0]), ((1 2), [1])
(c) ((1), [0]), ((1 2 3), [1]), ((1 3 2), [0]), ((1), [1]), ((1 2 3), [0]), ((1 3 2), [1])
5. Problem 16.2 on page 87. Determine the right cosets of h[3]iin Z12 .
{[0], [3], [6], [9]} {[1], [4], [7], [10]} {[2], [5], [8], [11]}