State the structure theorem for finitely generated Abelian groups.
Up to isomorphism, what are the Abelian groups of order \(300 = 2^2 \cdot 3 \cdot 5^2\)?
Solution
Write \(C_m=\mathbb Z/m\mathbb Z\).
1. The elementary-divisor form of the structure theorem is as follows.
Every finitely generated abelian group \(A\) has a decomposition \[ A\cong\mathbb Z^r\oplus \bigoplus_{p\ \mathrm{prime}} \bigoplus_{j=1}^{t_p} C_{p^{e_{p,j}}}, \] where \(r\) and the \(t_p\) are nonnegative integers, only finitely many \(t_p\) are nonzero, and \(1\leq e_{p,1}\leq\cdots\leq e_{p,t_p}\). The integer \(r\) and the lists of exponents for each prime are uniquely determined by \(A\). This is the structure theorem for finitely generated abelian groups [@DF04]. Empty sums are allowed.
2. The four groups of order \(300\) are the following.
| Group | Exponent |
|---|---|
| \(C_4\oplus C_3\oplus C_{25}\) | \(300\) |
| \(C_2\oplus C_2\oplus C_3\oplus C_{25}\) | \(150\) |
| \(C_4\oplus C_3\oplus C_5\oplus C_5\) | \(60\) |
| \(C_2\oplus C_2\oplus C_3\oplus C_5\oplus C_5\) | \(30\) |
1. For each prime \(p\), thesum \(\sum_j e_{p,j}\) is the exponent of \(p\) in the group order.For \(p=2\) that sum is \(2\), with exactly the partitions \((2)\)and \((1,1)\). For \(p=3\) it is \(1\), with only the partition \((1)\).For \(p=5\) it is \(2\), again with the partitions \((2)\) and \((1,1)\).No other prime occurs. The two independent choices for the\(2\)-part and the \(5\)-part give precisely the four displayed groups.Each has order \(4\cdot3\cdot25=300\).
The exponent of a finite abelian group is the least positive integer that annihilates every element. For a direct sum of cyclic groups, it is the least common multiple of their orders: an integer kills the whole group exactly when it kills a generator in every summand. This gives the four exponents displayed in the table. They are distinct and invariant under isomorphism, so no two listed groups are isomorphic. The primary-decomposition argument proves that every abelian group of order \(300\) appears on the list.