I was studying about a group of order 60, and I found this page; and it says

In the case of a finite nilpotent group (which in this case coincides with finite abelian group), the number of subgroups of a given order is the product of the number of subgroups of order equal to each of its maximal prime power divisors, in the corresponding Sylow subgroup.

but I can't even understand what it means. due to my bad English I guess

could you please explain what it means as simple as possible? and give a proof or hint for this statement?


Let's say our finite nilpotent group $G$ has order $n=p_1^{e_1}\cdots p_k^{e_k}$. Then $G$ is equal to a direct product $P_1\times\cdots\times P_k$ where $P_i$ is our Sylow $p_i$-subgroup.

Now, we are curious how many subgroups $G$ has of order $m=p_1^{d_1}\cdots p_k^{d_k}$. The statement says that if our Sylow subgroup $P_i$ has $r_i$ subgroups of order $p_i^{d_i}$, then $G$ has $r_1\cdots r_k$ subgroups of order $m$. This isn't difficult to see since we already mentioned that $G$ is a direct product of its Sylow subgroups.


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