Abstract algebra (another question about isomorphic and direct product )
[From: ] [author: ] [Date: 11-12-29] [Hit: ]
1]) then s = t mod mn, so s - t = kmn for some k, so s[1,1] - t[1,1] = (s-t)[1,1] = kmn[1,......
We now only need to show that f is a bijection. It is clear that the mn multiples of [1,1] map to all mn elements of Z_mn, so it must be 1-1 and onto. If f(s[1,1]) = f(t[1,1]) then s = t mod mn, so s - t = kmn for some k, so s[1,1] - t[1,1] = (s-t)[1,1] = kmn[1,1] = [0,0], so s[1,1] = t[1,1]
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Suppose that Z_m x Z_n is cyclic. Then it has a generator. Suppose this generator is (g, h). Then, clearly, g is a generator of Z_m and h is a generator of Z_n. Observe that for any k than is a multiple of m, (k+1) g = g (and a similar observation for h and n). Let L be the least common multiple of m and n. It follows that (L+1)(g,h) = ((L+1)g, (L+1)h) = (g,h). So, there can be at most L elements in the set {k (g,h) | k \in N}. Because (g,h) is a generator of Z_m x Z_n, this set must have exactly mn elements, so L >= mn. This forces L = mn and (m,n) = mn/L = 1.
Conversely, suppose (m,n) = 1. We show that Z_m x Z_n is cyclic. Let g and h be generators of Z_m and Z_n respectively. Consider the set S = {(kg, kh) | 0 <= k <= mn - 1}. We claim that this set has cardinality mn. Suppose not. Then there are distinct k1 and k2, both in the range [0, mn-1] such that k1 g = k2 g (in Z_m) and k1 h = k2 h (in Z_n). Let k1 > k2 (the case k2 > k1 is symmetric). This forces (k1 - k2) g = 0 and (k1 - k2) h = 0. Since g and h are generators, m divides (k1 - k2) and n divides (k1 - k2). Further, (m,n) = 1, so mn divides (k1 - k2). But, 0 < k1 - k2 <= mn - 1, so this is impossible. Contradiction. Hence, cardinality of S is mn. Consequently, it must be the entire set Z_m x Z_n. So, (g,h) is a generator of Z_m x Z_n.
The isomorphism follows from the converse result above and the observation that any two finite cyclic groups of the same size are isomorphic.
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Suppose d = (m, n). I claim that, for any (a, b) in Z_m x Z_n:
(a, b)^(mn/d) = (0, 0)
Notice that (x, y)^c = (cx, cy), since we are in an additive group. Thus, we need to prove that:
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