- Determine the elementary divisors and invariant factors of the Abelian group \(\mathbf{Z}_{15}\times\mathbf{Z}_{20}\times\mathbf{Z}_9\) of order 2700.
- Determine the number of nonisomorphic Abelian groups of order 2700.
Solution
Let \(G=\mathbf{Z}_{15}\times\mathbf{Z}_{20}\times\mathbf{Z}_9\).
1. The elementary divisors of \(G\) are \(2^2,3,3^2,5,5\), and its invariant factors are \(15\mid180\).
If \(\gcd(a,b)=1\), then \(\mathbf{Z}_a\times\mathbf{Z}_b\cong\mathbf{Z}_{ab}\). Hence \(\mathbf{Z}_{15}\cong\mathbf{Z}_3\times\mathbf{Z}_5\), \(\mathbf{Z}_{20}\cong\mathbf{Z}_{2^2}\times\mathbf{Z}_5\), and \[ G\cong\mathbf{Z}_{2^2}\times\mathbf{Z}_3\times\mathbf{Z}_{3^2}\times\mathbf{Z}_5\times\mathbf{Z}_5, \] which lists the elementary divisors. Grouping the largest prime powers gives \(\mathbf{Z}_{2^2}\times\mathbf{Z}_{3^2}\times\mathbf{Z}_5\cong\mathbf{Z}_{180}\), and the rest gives \(\mathbf{Z}_3\times\mathbf{Z}_5\cong\mathbf{Z}_{15}\). Thus \(G\cong\mathbf{Z}_{15}\times\mathbf{Z}_{180}\) with \(15\mid180\), and by uniqueness of invariant factors [@DF04] these are the invariant factors of \(G\).
2. There are \(\boxed{12}\) nonisomorphic abelian groups of order \(2700\).
Since \(2700=2^2\cdot3^3\cdot5^2\), an abelian group of order \(2700\) is determined up to isomorphism by one partition of each exponent, the parts giving the orders of its cyclic primary factors [@DF04]. The exponent \(2\) has the \(2\) partitions \((2),(1,1)\), and the exponent \(3\) has the \(3\) partitions \((3),(2,1),(1,1,1)\). The number of groups is therefore \(2\cdot3\cdot2=12\).
3. Q.E.D.
Step 1 answers part (a) and step 2 answers part (b).