# Download A Hungerford’s Algebra Solutions Manual by James Wilson PDF

By James Wilson

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Notice a/pn − a(1/pn ) = 0 ∈ Z, so a/pn = a1/pn ; therefore, Z(p∞ ) = 1/pn | n ∈ Z+ . Notice that given any m > n, pm−n 1/pm = 1/pn so any set {1/pn | n ∈ Z+ , n > k} is also a generating set for any k ∈ Z+ . 17 Join of Abelian Groups. Let G be an abelian group and let H,K be subgroups of G. Show that the join H ∨ K is the set {ab | a ∈ H, b ∈ K}. Extend this result to any finite number of subgroups of G. Proof: Denote the set {ab | a ∈ H, b ∈ K} by HK. Since e ∈ H and e ∈ K, the elements ae = a and eb = b are in HK for all a ∈ H, b ∈ K.

Again (1234)5 = (1234) and clearly this implies (1234)3 = (1234)−1 so these elements are in fact all the elements of our subgroup generated by (1234). Define the mapping f : Z4 → i by f (1) = i; the mapping g : Z4 → (1234) by f (1) = (1234). 2 these functions determine a unique homomorphism. By the Principle of refinement f (Z4 ) and g(Z4 ) are subgroups and as we have shown both are surjective. We furthermore have a natural inverse mapping f 1− (i) = 1 and g −1 ((1234)) = 1. 2. Finally f f −1 (in ) = in , so f f −1 = 1 i ; f −1 f (n · 1) = n · 1 so f −1 f = 1Z4 ; gg −1 ((1234)n ) = (1234)n , so gg −1 = 1 (1234) ; g −1 g(n · 1) = n · 1 thus g −1 g = 1Z4 .

But we notice any primes removed from m m by (m, n) are replaced by their copy in n, that is, [n, (m,n) ] = [n, m]; thus, k = [m, n] and so we have (m, n)a + b as our element of order [m, n]. 3. 4. Use Theorem- Hint(1/5): It is best simply to multiply and observe. Zpq . Let G be an abelian group of order pq, with (p, q) = 1. Assume there exists a, b ∈ G such that |a| = p, |b| = q and show that G is cyclic. 3 there must exist an element of order [p, q]. As before we know pq = [p, q](p, q) which is simply pq = [p, q] in our case.