Give the lists of elementary divisors for an abelian group of the order $270$ and match each list with corresponding list of invariant factors.
Why the elementary divisors corresponding to invariant factor $270$ is $2, 27,5$, not $2,9,3,5$ or $2,3,3,3,5$? And why the elementary divisors corresponding to invariant factor $90,3$ is $2, 9,3,5$ and the elementary divisors corresponding to invariant factor $30,3,3$ is $2, 3,3,3,5$?
Best Answer
From the fundamental theorem of abelian groups, there can be exactly $P(n)$ number of non-isomorphic group of order $p^n$, where $P(n)$ is the number of integer partitions of $n$.
For $n=\prod_{i=0}^n p_i^{n_i}$, where each $p_i$'s are distinct primes, we have $\mathbb{Z}_n=\oplus_{i=0}^n \mathbb{Z}_{p_i^{n_i}}$. Say $A_i$ is the set of all non-isomorphic groups of order $p_i^{n_i}$. Then all the non-isomorphic groups of order $n$ is obtained by taking one group from each $A_i$ and taking their direct sums (note that the order of the direct sums doesn't matter). Thus the total number of non-isomorphic groups of order $n$ is $\prod_{i=0}^nP(n_i)$, $P(n_i)$ being the number of integer partitions of $n_i$
Now, $270 = 2.3^3.5$
The integer partitions of 3 are
So number of non-isomorphic groups of order $270$ is $P(1).P(3).P(1)=1.3.1=3$. The elementary factors : $(2, 3^3, 5), (2, 3^2, 3, 5), (2, 3, 3, 3, 5)$ i.e any abelian group of order $270$ is isomorphic to one of the following groups :
Now for the invariant factors, using the technique described here, we have the invariant factors : $(270), (3, 90), (3, 3, 30)$
Thus the list of non-isomorphic groups :