By Randall R. Holmes

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We denote by gcd(m, n) the greatest common divisor of m and n. For instance gcd(12, 18) = 6. If gcd(m, n) = 1, the integers m and n have no prime factors in common and they are said to be relatively prime. Looking at the computations in the example, we see that the two components are out of sync, reaching 0 at different steps. They get together only at the step 6 = (2)(3). This occurs because 2 and 3 are relatively prime. 2 Theorem. Zm ⊕ Zn ∼ = Zmn if and only if gcd(m, n) = 1. Proof. (⇒) Assume that Zm ⊕ Zn ∼ = Zmn .

To show that the two functions στ and τ σ are equal, we need to show that (στ )(m) = (τ σ)(m) for all m in their common domain, which is {1, 2, . . , n}. Let m ∈ {1, 2, . . , n}. First assume that m ∈ / I ∪ K (in other words, assume that m does not appear in either cycle). Then σ and τ both fix m giving (στ )(m) = σ(τ (m)) = σ(m) = m = τ (m) = τ (σ(m)) = (τ σ)(m). Now assume that m ∈ I. Then σ(m) ∈ I, as well. Since I and K are disjoint by assumption, m and σ(m) are not in K, so they are fixed by τ .

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. However, this is not a valid argument for showing that Q Z since it uses order properties of numbers and not just the binary operations involved.