Orders of elements in finite abelian groups (D&F 5.2.5)

Let \(G\) be a finite abelian group of type \((n_1,n_2,\ldots,n_t)\). Prove that \(G\) contains an element of order \(m\) if and only if \(m \mid n_1\). Deduce that \(G\) is of exponent \(n_1\).

Let \(x = (x_1,x_2,\ldots,x_t) \in G\) have order \(m\). By defining \(a_i\) to be the \(t\)-tuple with \(x_i\) in its \(i\) component and 0 elsewhere, we can write \(x\) as a product of \(a_i\)'s. For all \(i\), we have that \(\left|a_i\right| \mid n_i \mid n_1\). Thus \(m\), which is precisely the LCM of all \(\left|a_i\right|\)'s, divides \(n_1\).

Conversely, if \(m \mid n_1\), then by virtue of being cyclic we can find an element \(a \in Z_{n_1}\) with order \(m\). Then the element \((a,1,\ldots,1)\) clearly has order \(m\) in \(G\). The statement that \(G\) has exponent \(n_1\) follows immediately from the fact that \(G\) has an order \(n_1\) element (namely \((b,1,\ldots,1)\) where \(Z_{n_1} = \langle b \rangle\)), and that orders of any element must divide \(n_1\).