How many mutually non-isomorphic Abelian groups whose order is $3^{2}*11^{4}*7$ are there?
Can anyone give a quick (obviously logical) solution?
How many mutually non-isomorphic Abelian groups whose order is $3^{2}*11^{4}*7$ are there?
Can anyone give a quick (obviously logical) solution?
The answer is $\,2\cdot 5=10\,$, but way more important and interesting is how to reach that answer, which is based in the Fundamental theorem for finitely generated abelian groups.
Definition: A partition of a number $\,n\in\Bbb N:=\{1,2,...\}\,$ is a finite sum of the form
$a_1+a_2+...+a_r=n\,\,\,\,,\,\,a_i\in\Bbb N\;\;\;\forall\,r$
We say two partitions of the same natural number are identical if they only difer in the order of the summands, and let us define $\,\cal P(n):=\,$number of different partitions of $\,n\,$
Theorem: If $\,n=p_1^{n_1}\cdot\ldots\cdot p_k^{n_k}\,$ is the prime decomposition of the natural number $\,n\,$ then there are $\,\prod_{m=1}^k\cal P(n_k)\,$ different abelian groups of order $\,n\,$ up to isomorphism