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Abstract Algebra I by Randall R. Holmes

By Randall R. Holmes

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6 Example addition). Is Q isomorphic to Z? (both viewed as groups under Solution It was pointed out in Section 2 that |Q| = |Z|, which is to say that 36 there exists a bijection from Q to Z. 3, that Q Z. Instead, we might try to imagine some property that Q has that Z does not have. We observe that between any two distinct elements x and y of Q there exists another element of Q (namely, (x + y)/2). But this is not the case for Z since, for instance, there is no integer between the integers 1 and 2.

This permutation is called an r-cycle (or a cycle of length r) and we write length(σ) = r. A cycle is unchanged if the last number is moved to the first. For instance: (1, 5, 2, 4) = (4, 1, 5, 2) = (2, 4, 1, 5) = (5, 2, 4, 1). ” If the numbers are arranged in order around a circle, then a cyclic permutation corresponds to a rotation of the circle. 59 The inverse of a cycle is obtained by writing the entries in reverse order. For example, (1, 5, 2, 4)−1 = (4, 2, 5, 1). A transposition is a 2-cycle.

Since gcd(m, n) = 1, it follows that mn divides k. In particular, mn ≤ k. On the other hand k is the order of the cyclic subgroup of Zm ⊕ Zn generated by (1, 1), so k is less than or equal to the order of Zm ⊕ Zn , which is mn. We conclude that k = mn, so in fact (1, 1) = Zm ⊕ Zn . Therefore Zm ⊕ Zn is cyclic. 1, Zm ⊕ Zn ∼ = Zmn . 57 6 – Exercises 6–1 Let G be a cyclic group of order n. Prove that if d is a positive integer dividing n, then G has a subgroup of order d. 1. First work with the special case G = Z12 to get an idea for the proof.

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