SLIDE 3 Finite abelian groups
Proposition 1
Znm ∼ = Zn × Zm if and only if gcd(n, m) = 1.
Proof (sketch)
“⇐”: Suppose gcd(n, m) = 1. We claim that (1, 1) ∈ Zn × Zm has order nm. To prove the claim, let k denote the order of the element (1, 1) ∈ Zn × Zm. Then (k, k) = (0, 0). This means n | k and m | k. In fact, k is lcm(n, m) the smallest common multiple of n and m. Since n and m has no common divisor, lcm(n, m) = nm. So k = nm.
(1,0) (2,0) (3,0) (0,1) (1,1) (2,1) (3,1) (0,2) (1,2) (2,2) (3,2)
· · ·
(0,0) (1,1) (2,2) (3,0) (0,1) (1,2) (2,0) (3,1) (0,2) (1,0) (2,1) (3,2)
Z4 × Z3 ∼ = Z12
Sec 4.4 Finitely generated abelian groups Abstract Algebra I 3 / 7