By James Wilson

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**Sample text**

Suppose a/b is not in lowest terms so that there exists an a and b such that a/b = a /b and a /b is in lowest terms. Then (b, b ) = b and thus (b , p) = (b, b , p) = 1 so the selection of fractions with denominators relatively prime to p is well-defined. Therefore Rp is given by a well-defined rule and thus is defined. Furthermore any integer k has denominator 1; (1, p) = 1, so Z ⊂ Rp ; thus 0 is in Rp . Since 0 is the additive identity of Q and we adopt the same addition in Rp it is clear 0 is the identity of Rp – if Rp is shown to have a well-defined addition.

Now ab = 1 0 Example: Take the elements (1, 1) and (0, −1) in Z2 ⊕ Z. Given any integer n, n(1, 1) = (n · 1, n), so n(1, 1) = m(1, 1) only if m = n, and likewise n(0, −1) = m(0, −1) only when −n = −m, or simply m = n. Therefore both elements generate infinite cyclic groups, so they have infinite order by definition. However (1, 1) + (0, −1) = (1, 0) which clearly has order 2 since 2(1, 0) = (0, 0). 6 Cyclic Elements. If G is a cyclic group of order n and k|n, then G has exactly one subgroup of order k.

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. 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.