By Randall R. Holmes

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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 . Therefore, (στ )(m) = σ(τ (m)) = σ(m) = τ (σ(m)) = (τ σ)(m). Similarly, if m ∈ K, then (στ )(m) = (τ σ)(m). We have shown that in all cases (στ )(m) = (τ σ)(m). Since m was an arbitrary element of {1, 2, .

32 G1 × G2 is the direct product of the groups G1 and G2 . , the binary operations are both +), then the direct product is called the direct sum and it is denoted G1 ⊕ G2 . In this case, the operation is denoted + and it is called componentwise addition: (x1 , x2 ) + (y1 , y2 ) = (x1 + y1 , x2 + y2 ). 10 Isomorphism Consider the group Z = {. . , −2, −1, 0, 1, 2, . . } under addition. If we create a new group Z = {. . , −2, −1, 0, 1, 2, . . } by putting bars over each element of Z and by using a binary operation that acts just like + acts on the unadorned integers (so, for instance, 2 3 = 5), then the group (Z, ) is essentially the same as the original group (Z, +).

Then m = km and n = kn for some integers m and n . We have km n (a, b) = (km n a, km n b) = (n ma, m nb) = (0, 0). Since (a, b) has order mn and km n > 0 it follows that mn ≤ km n . So mn ≤ km n ≤ km kn = mn. Since the ends of this string are the same, the intermediate ≤ signs must actually be equalities. Thus km n = km kn which implies k = 1 as desired. (⇐) Assume that gcd(m, n) = 1. Let k be the order of the element (1, 1) of Zm ⊕ Zn . Then (k1, k1) = k(1, 1) = (0, 0). 3, m divides k and n divides k.