How many abelian groups are there of order 36?
Solution
Since \[ 36=2^2 3^2, \] the \(2\)-primary part of an abelian group of order \(36\) is either \(C_4\) or \(C_2\oplus C_2\), and independently the \(3\)-primary part is either \(C_9\) or \(C_3\oplus C_3\).
1. There are exactly four isomorphism classes.
The partitions of \(2\) are \(2\) and \(1+1\). Thus there are two choices for the \(2\)-primary component and two choices for the \(3\)-primary component. Primary decomposition is unique, so the total number is \[ 2\cdot2=4. \]
2. In invariant-factor form the four groups are\[ C_{36},\qquad C_3\oplus C_{12},\qquad C_2\oplus C_{18},\qquad C_6\oplus C_6. \]
Combine the primary possibilities using the Chinese remainder theorem: \[ C_4\oplus C_9\cong C_{36}, \] \[ C_4\oplus C_3\oplus C_3\cong C_3\oplus C_{12}, \] \[ C_2\oplus C_2\oplus C_9\cong C_2\oplus C_{18}, \] and \[ C_2\oplus C_2\oplus C_3\oplus C_3\cong C_6\oplus C_6. \] These invariant-factor lists are pairwise distinct, hence so are the groups.