Last Update : May 26, 1998

ALGEBRA I 1998 TAKE-HOME MIDTERM SOLUTION

Problem 1

  1. Since H is not empty, there is an element a in H. By the assumption, 1 = a-1a is also an element of H. Let a be an arbitrary element of H. Since 1 is an element of H, a-1=a-11 is also an element of H. Let a and b be arbitrary elements of H. Since we have already shown that a-1 is an element of H, ab = (a-1)-1b is also an element of H. Therefore H is a subgroup of G.
  2. Suppose ab-1 is an element of H. Since H is a subgroup, Hab-1 is a subset of H. On the other hand, since (ab-1)-1 is also an element of H, H = H1 = H(ab-1)-1ab-1 is a subset of Hab-1. Therefore, Hab-1 = H. By multiplying b from the right, we have Ha = Hb.
    Conversely, assume Ha = Hb. By multiplying b-1 from the right, we have Hab-1 = H. Since 1 is an element of H, ab-1=1ab-1 belongs to H as desired.

Problem 2

  1. The orders are 8, 8, and 16 respectively.
  2. Since the operation here is the sum, n-th power of 1 is the residue of n by 16, hence every element of Z16 appears in the subgroup generated by 1. Therefore Z16 is cyclic.
  3. Since Z16 is a cyclic group of order 16, for each divisor n of 16, there is a unique subgroup of Z16 generated by 16/n. Thus the subgroups are cyclic groups generated by 0, 1, 2, 4, 8 of order 1, 16, 8, 4, 2 respectively.

Problem 3

  1. 1, 3, 5, 7, 9, 11, 13, 15
  2. 72 = 92 = 1. Hence there are two subgroup of order 2. Thus it is not cyclic.
  3. {1}, {1, 7}, {1, 9}, {1,15}, {1, 7, 9, 15}, {1, 3, 9, 11}, {1, 5, 9, 13}, and Z*16.

Problem 4

  1. If X and Y belongs to O(3,R), then
    XY(XY)t = XYYtXt = XXt = I. Hence XY belongs to O(3,R). Since X-1 = Xt and XXt = XtX = I, X-1 also belongs to O(3, R) as well. Hence O(3,R) is a subgroup of GL(3,R).
  2. If X and Y belongs to G, then
    XYC = C and X-1C = C, G is a subgroup of O(3,R).
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