By James Wilson

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**Extra resources for A Hungerford’s Algebra Solutions Manual **

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Infinite Cyclic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 47 48 48 48 49 49 51 51 52 Order of Elements. 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.

It may be useful to know mn = (m, n)[m, n], where [m, n] is the least common multiple of m and n. 2 Orders in Abelian Groups. Let G be an abelian group containing elements a and b of orders m and n respectively. Show that G contains an element whose order is the least common multiple of m and n. 1 for our abelian group, [m, n]((m, n)a + b) = [m, n](m, n)a + [m, n]b = (mn)a + [m, n]b = 0 + 0 = 0; therefore, the order, k, of (m, n)a + b divides [m, n]. We are assuming: k((m, n)a + b) = k(m, n)a + kb = 0.

Proof: Adopt an additive notation throughout the proof. The zero element has order 1 trivially; thus T is never empty. Given two torsion elements a, b ∈ T , with |a| = m and |b| = n, consider their sum. 1 part (iv) which is where abelian comes into play) we see that a + b is a torsion element, a + b ∈ T , so T is closed. 3 we know |a| = | − a|, so T is closed to inverses. 5. 2. 10 Infinite Cyclic Groups. An infinite group is cyclic if and only if it is isomorphic to each of its proper subgroups.