Determine (up to isomorphism) all abelian groups of order 64; do the same for order 96.
Solution
By the classification of finite abelian groups, a finite abelian group is the direct sum of its Sylow subgroups, and an abelian group of order \(p^n\) is determined up to isomorphism by the partition \(n=a_1+\cdots+a_r\) with \(G\cong\ZZ_{p^{a_1}}\oplus\cdots\oplus\ZZ_{p^{a_r}}\).
1. There are exactly eleven abelian groups of order \(64=2^6\):\[ \begin{aligned} &\ZZ_{64},\\ &\ZZ_{32}\oplus\ZZ_2,\\ &\ZZ_{16}\oplus\ZZ_4,\\ &\ZZ_{16}\oplus\ZZ_2\oplus\ZZ_2,\\ &\ZZ_8\oplus\ZZ_8,\\ &\ZZ_8\oplus\ZZ_4\oplus\ZZ_2,\\ &\ZZ_8\oplus\ZZ_2\oplus\ZZ_2\oplus\ZZ_2,\\ &\ZZ_4\oplus\ZZ_4\oplus\ZZ_4,\\ &\ZZ_4\oplus\ZZ_4\oplus\ZZ_2\oplus\ZZ_2,\\ &\ZZ_4\oplus\ZZ_2\oplus\ZZ_2\oplus\ZZ_2\oplus\ZZ_2,\\ &\ZZ_2^{\oplus6}. \end{aligned} \]
An abelian group of order \(2^6\) is uniquely determined by a partition of \(6\): \[ 6,\ 5+1,\ 4+2,\ 4+1+1,\ 3+3,\ 3+2+1,\ 3+1+1+1, \] \[ 2+2+2,\ 2+2+1+1,\ 2+1+1+1+1,\ 1+1+1+1+1+1. \] Replacing each part \(a\) by a cyclic summand \(\ZZ_{2^a}\) gives exactly the displayed list. Distinct partitions give nonisomorphic groups by uniqueness in the classification theorem.
2. There are exactly seven abelian groups of order \(96=2^5\cdot3\):\[ \begin{aligned} &\ZZ_{32}\oplus\ZZ_3,\\ &\ZZ_{16}\oplus\ZZ_2\oplus\ZZ_3,\\ &\ZZ_8\oplus\ZZ_4\oplus\ZZ_3,\\ &\ZZ_8\oplus\ZZ_2\oplus\ZZ_2\oplus\ZZ_3,\\ &\ZZ_4\oplus\ZZ_4\oplus\ZZ_2\oplus\ZZ_3,\\ &\ZZ_4\oplus\ZZ_2\oplus\ZZ_2\oplus\ZZ_2\oplus\ZZ_3,\\ &\ZZ_2^{\oplus5}\oplus\ZZ_3. \end{aligned} \]
The Sylow \(3\)-subgroup has order \(3\), hence is necessarily \(\ZZ_3\). The Sylow \(2\)-subgroup has order \(2^5\), and the partitions \[ 5,\ 4+1,\ 3+2,\ 3+1+1,\ 2+2+1,\ 2+1+1+1,\ 1+1+1+1+1 \] give its seven possible isomorphism types. Taking the direct sum with \(\ZZ_3\) gives the displayed list. Uniqueness of the primary decomposition shows that no two groups on the list are isomorphic and that no further cases occur.