A Hungerford’s Algebra Solutions Manual by James Wilson

By James Wilson

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Extra resources for A Hungerford’s Algebra Solutions Manual

Example text

Join of Abelian Groups . . . Join of Groups . . . . . . Subgroup Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 37 37 37 38 38 39 39 40 40 40 41 41 42 42 44 44 44 45 Homomorphisms.

Clearly an+1 is defined; however, it must also lie in a which has only n representatives; thus, there exists a j ≤ n such that an+1 = aj . If j = 1 then an+1 = a. Suppose j > 1, then aj−1 is defined and still n + 1 > j, so an+1−j+1 is defined and so an+1−j+1 aj−1 = an+1 = aj = aaj−1 . By cancellation we see an−j+2 = a. Therefore for some power 2 ≤ k ≤ n, ak = a, (k = n−j+2), and ak−1 is defined, and combined with the case j = 1 we see ak = a for some 1 < k ≤ n + 1 and ak−1 is also defined.. Now take any b ∈ G.

7 this set generates Z ⊕ Z. Additionally Z ⊕ Z cannot be generated by less than 2 elements. 8 claims it generates only {n(a, b) | n ∈ Z}. However n(a, b) = (na, nb). Yet (−a, b) is clearly also an element of Z ⊕ Z. If it is not in (a, b) then by definition the Z ⊕ Z is not generated by one element. If however (−a, b) ∈ (a, b) , then (−a, b) = (na, nb) requires b = nb, or simply n = 1. Thus −a = a so a = 0. Therefore the element (1, 0) is excluded from (a, b) . Therefore no lone element generates Z ⊕ Z.

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