A divisibility criterion for orders of elements in a finite abelian group

Let G be a finite abelian group of type (n_1,n_2,\ldots,n_k). (That is, n_1,\ldots,n_k are the invariant factors of G.) Prove that G contains an element of order m if and only if m|n_1. Deduce that G is of exponent n_1.


(\Rightarrow) Suppose G contains an element x of order m. Now x = (a_1,a_2,\ldots,a_k) for some a_i \in Z_{n_i}, and we have m = \mathsf{lcm}(|a_1|,|a_2|,\ldots,|a_k|). By the divisibility criterion for invariant factors, n_1 is a common multiple of the n_i, hence of the |a_i|. Thus m|n_1.

(\Leftarrow) Suppose m|n_1. Now G has a subgroup isomorphic to Z_{n_1}, which (being cyclic) has an element of order m. Thus G has an element of order m.

We conclude that if x \in G has order m, then since n_1 = tm for some t, n_1x = tmx = 0. Hence the exponent of G is finite and divides n_1. Moreover, the exponent of G is at least n_1 since G has an element of order n_1. Thus G has exponent n_1.

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