Practice Final Examination Math 100B Solutions

We note that the Chinese remainder theorem implies that this group is isomorphic with the cylic group of order 30, Z30. 31 is prime so the only abelian group of order 31 is Z31. 32 = 2. So there are many possibilities corresponding to all of the ways we can write m1 +m2 + ...+mr = 5 with m1 ≥ m2 ≥ ... ≥ mr. The solutions are 5 = 5 which corresponds to Z32, 4 ≥ 1 giving Z16 ×Z2, 3 ≥ 2 giving Z8 ×Z4, 3 ≥ 1 ≥ 1 giving Z8 ×Z2 ×Z2„ 2 ≥ 2 ≥ 1, giving Z4×Z4×Z2, 2 ≥ 1 ≥ 1 ≥ 1 giving Z4 × Z2 × Z2 × Z2 and 1 ≥ 1 ≥ 1 ≥ 1 ≥ 1, giving Z2 × Z2 × Z2 × Z2 × Z2. We will now show that these groups are pairwise non-isomorphic. We first observe that Z32 contains an element of order 32 and no other on the list does. Z16×Z2 contains an element of order 16 and the only other group listed that has such an element is Z32 which has been excluded. We are left with Z8 × Z4,Z8 × Z2 × Z2, Z4 × Z4 × Z2, Z4 × Z2 × Z2 × Z2, Z2 × Z2 × Z2 × Z2 × Z2. Every element in the last has order 2 and every other has at least one element of order 4. The first 2 have elements of order 8 and the next 2 do not. Thus the only possible isomorphisms are between Z8 × Z4and Z8 ×Z2 ×Z2 and between Z4 ×Z4 ×Z2,and Z4 ×Z2 ×Z2 ×Z2. Suppose that φ : Z8×Z4 → Z8×Z2×Z2 is an isomorphism (onto). Then φ(2(x, y)) = 2φ(x, y) for (x, y) ∈ Z8 × Z4. Now 2Z8 = {[0], 2[1], 2[3], 2[4], 2[5], 2[6], 2[7]}