Classification of finite abelian groups

For a given order \(n\), factor \(n=p_1^{a_1}\cdots p_s^{a_s}\). Every abelian group of order \(n\) splits into one component per prime, of order \(p_i^{a_i}\), and that component is a product of cyclic groups \(\mathbb{Z}_{p_i^{k}}\) whose exponents \(k\) add up to \(a_i\). Hence abelian groups of order \(p^a\) correspond one to one with the partitions of the integer \(a\), and the total for \(n\) is the product of the partition numbers of the \(a_i\).

Guiding example: for \(n=12=2^2\cdot3\) the partitions of \(2\) are \(2\) and \(1+1\), and the one of \(1\) is unique. This gives \(\mathbb{Z}_4\times\mathbb{Z}_3\cong\mathbb{Z}_{12}\) and \(\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_3\cong\mathbb{Z}_2\times\mathbb{Z}_6\). If every \(a_i=1\) (squarefree order), there is only one group and it is cyclic.

Uniqueness means that two different products of this kind are never isomorphic: the list of factors \(\mathbb{Z}_{p^k}\) is determined by \(G\). A typical mistake is to believe that \(\mathbb{Z}_2\times\mathbb{Z}_4\) and \(\mathbb{Z}_8\) coincide because they have the same order; the first has no element of order \(8\). An equivalent presentation uses invariant factors \(d_1\mid d_2\mid\cdots\mid d_m\) with \(G\cong\mathbb{Z}_{d_1}\times\cdots\times\mathbb{Z}_{d_m}\).

The proof uses that a direct product of cyclic groups of coprime orders is cyclic (Chinese remainder theorem), and the result is the special case, for \(\mathbb{Z}\)-modules, of the structure theorem over principal ideal domains. It contrasts with non-abelian groups, for which no such simple list exists.