New PDF release: A Hungerford’s Algebra Solutions Manual

By James Wilson

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So A2 = B 2 and −I 2 = (−1)2 II = 1I = I; thus A4 = B 4 = I and moreover A3 = −IA = −A and B 3 = −B = −IB = A2 B. Next Let P = PA = P 0 1 −1 0 = −1 0 0 1 = 0 1 −1 P = A3 P ; 0 therefore, BA = iP A = iA3 P = A3 iP = A3 B. Here we see Q8 will not be abelian. Multiplication in Q8 has a normal form: given any product of A’s and B’s we may express it in the form Ai B j for some integers i and j; moreover, B 3 = −B = −IB = A2 B so we in fact need only elements of the form Ai and Ai B. Since A has order 4 we have at least the following elements: Q8 = {I, A, A2 , A3 , B, AB, A2 B, A3 B} and if we write these as matrices we see these are all distinct elements: Q8 = 1 0 0 0 1 −1 , , 1 −1 0 0 0 0 , −1 1 −1 0 , 0 i i i , 0 0 0 0 −i −i , , −i −i 0 0 Notice for all X ∈ Q8 , X = I, −I, X −1 = −X; thus the group is often described as ˆ −k}, ˆ Q8 = {ˆ1, −ˆ1, ˆi, −ˆi, ˆj, −ˆj, k, where ˆi = A, ˆj = B, kˆ = AB.

2 we proved a/pi = a1/pi . In part (e) notice G is defined in terms of equivalence classes so the case for well-defined must be explicit. Groups 50 (e) Let x1 , x2 , . . be elements of an abelian group G such that |x1 | = p, px2 = x1 , px3 = x2 , . . , pxn+1 = xn , . . The subgroup generated by the xi (i ≥ 1) is isomorphic to Z(p∞ ). ] Proof: (a) Every element in Z(p∞ ) is of the form a/pi for some i ∈ N. In taking the sum of a/pi , pi times we have a/pi +· · ·+a/pi = pi (a/pi ) = a ∼ 0; therefore, (pi )a/pi = 0.

Since the set is finite, am ≡ an (mod p) for some m and n, m = n, or otherwise there would be infinitely many elements. Without loss of generality let m < n. 8, part (iii), and the fact we know (a, p) = 1 so (am , p) = 1, we conclude 1 ≡ an−m (mod p). Now certainly this requires ak ≡ 1 (mod p) for some positive integer k, and we take the least such k. Therefore either k = 1 which implies a = 1 or ak−1 = 1 and aak−1 = ak−1 a = ak ≡ 1 (mod p) and so ak−1 = a−1 . Since Z∗p is closed to products, ak−1 ∈ Z∗p .

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A Hungerford’s Algebra Solutions Manual by James Wilson


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