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Additional resources for A Hungerford’s Algebra Solutions Manual

Sample text

11 41 Center. If G is a group, then C = {a ∈ G | ax = xa for all x ∈ G} is an abelian subgroup of G. C is called the center of G. Proof: The identity element has the property ex = x = xe for any element x ∈ G; thus e ∈ C so C is nonempty. Given any two elements a, b ∈ C, xa = ax and xb = bx for all x ∈ G, which allows x(ab) = (xa)b = (ax)b = a(xb) = a(bx) = (ab)x. So ab ∈ C. Finally xa = ax implies a−1 x = a−1 x(aa−1 ) = a−1 (xa)a−1 = (a−1 a)xa−1 = xa−1 . Therefore C is closed to inverses so it is a subgroup of G.

Let a, b be elements of a group G. Show that |a| = |a−1 |; |ab| = |ba|, and |a| = |cac−1 | for all c ∈ G. Proof: Consider the cyclic group generated by an element a. 3, |a| = |a−1 |. Suppose the order, n, of ab is finite, so that (ab)n = e. We re-associate the product as follows: (ab) · · · (ab) = a(ba) · · · (ba)b = a(ba)n−1 b. So a(ba)n−1 b = e which implies (ba)n−1 = a−1 b−1 = (ba)−1 , and thus finally (ba)n = e. Therefore the order of ba is less than or equal to the order of ab. However the argument is completely symmetric if we begin with the order of ba (notice this has implicitly guaranteed the order of ba is finite so we satisfy the hypothesis to begin); thus we have |ba| ≤ |ab| and |ab| ≤ |ba| so by the antisymmetry of integer ordering we know |ab| = |ba|.

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. Thus HK contains H and K. 8 that every element in the join is of the form an1 1 · · · ant t , with ai ∈ H ∪ K, ni ∈ Z. But we know G to be abelian so we may commute all the elements so that we begin with all the elements in H and end with the elements ni nij+1 n n in K.