Let p1,......,pn be distinct primes. Up to isomorphism ,how many Abelian groups are there of order p14p24....pn4

$ Attempt$

Each pi4 can be partitioned into 5 ways. So total number of Abelian groups is 5n

Is my approach correct ?

  • 2
    $\begingroup$ Yes${}{}{}{}{}$. $\endgroup$ – Lord Shark the Unknown Jul 8 '18 at 6:36
  • $\begingroup$ Thanks......... $\endgroup$ – blue boy Jul 9 '18 at 3:33

Consider $2^4$... You can have $\mathbb Z_{2^4},\mathbb Z_2×\mathbb Z_{2^3},\mathbb Z_2×\mathbb Z_2×\mathbb Z_{2^2},\mathbb Z_{2^2}×\mathbb Z_{2^2}$, or $\mathbb Z_2×\mathbb Z_2×\mathbb Z_2×\mathbb Z_2$... Thus you have $5$ different groups of order $2^4$.

Yes, the number of partitions of $4$ is apparently $5$: $4,1+3,1+1+2,2+2$ and $1+1+1+1$.

So you have $5^n$ such abelian groups.


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