Let \(G\) be a finite abelian group with \(G\cong\bigoplus_{i=1}^s \ZZ/m_i\ZZ\) for integers \(m_i\geq 2\), not necessarily distinct.
The elementary divisors of \(G\) are the prime powers \(p^e\), \(e\geq1\), occurring in the factorizations \(m_i=\prod_p p^{e_{p,i}}\), listed with multiplicity over all \(i\); that is, \(G\cong\bigoplus_{i}\bigoplus_{p} \ZZ/p^{e_{p,i}}\ZZ\).
For each prime \(p\), list the exponents of the \(p\)-power elementary divisors in decreasing order, \(e_{p,1}\geq e_{p,2}\geq\cdots\), padding with zeros. Put \(d_j\coloneqq\prod_p p^{e_{p,j}}\) and let \(t\) be the largest \(j\) with \(d_j\neq 1\). Then \(d_t\divides d_{t-1}\divides\cdots\divides d_1\) and \[ G\cong \ZZ/d_t\ZZ\oplus\cdots\oplus\ZZ/d_1\ZZ \] is the invariant factor decomposition of \(G\).
By the Chinese remainder theorem, \(\ZZ/m\ZZ\cong\bigoplus_p\ZZ/p^{e_p}\ZZ\) for \(m=\prod_p p^{e_p}\), which gives (a). For (b), the same theorem gives \(\ZZ/d_j\ZZ\cong\bigoplus_p\ZZ/p^{e_{p,j}}\ZZ\), so \(\bigoplus_j\ZZ/d_j\ZZ\) has the same elementary divisors as \(G\) and is isomorphic to \(G\). Since \(e_{p,j+1}\leq e_{p,j}\) for every \(p\), \(d_{j+1}\divides d_j\).
Let \(G=\ZZ/12\ZZ\oplus\ZZ/18\ZZ\oplus\ZZ/5\ZZ\). Since \(12=2^2\cdot3\) and \(18=2\cdot3^2\), the elementary divisors are \(2^2, 2, 3^2, 3, 5\). The largest power of each prime gives \(d_1=2^2\cdot3^2\cdot5=180\), and the remaining powers give \(d_2=2\cdot3=6\). Hence \(G\cong\ZZ/6\ZZ\oplus\ZZ/180\ZZ\), with \(6\divides180\).