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Given an abelian group $G$ of order $N$, I want to know how many subgroups of order $n$ there are.

For example: let $G$ be an abelian group of order $72$. How many subgroups or order $8$ does $G$ have? And of order $4$?

I know that $8$ divides $72$, so there is at least one subgroup of order $8$ of $G$. But how can I know exactly how many there are?

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    In general, the number of subgroups with a given number of elements depends on the chosen group $G$ – 2017-01-12
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    Since $G$ is abelian, its Sylow subgroups are normal, hence there is only one of them per prime dividing the order of $G$. So there's one subgroup of order $8$ in an abelian group of order $72$. How many subgroups of order $4$ and $2$ there are depends on the structure of the $2$-Sylow subgroup. – 2017-01-12

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As Peter stated in his comment, this is with your knowledge impossible to say (except some few special cases). An easy example would be $N=4$ and $n=2$:

$\mathbb{Z} /4 \mathbb{Z}:$ has 1 such subgroup

$\mathbb{Z} /2 \mathbb{Z} \times \mathbb{Z} /2 \mathbb{Z}$: has 2 such subgroups

But if $(N/n,n)=1$ holds you know there is only 1 such subgroup (you can prove this with prime potences first).