
#1
Nov512, 01:15 PM

P: 126

I was wondering if there are any theorems that specify an exact number of subgroups that a group G has, maybe given certain conditions.
The closest thing I know is a theorem that says if G is finite and cyclic of order n it has exactly one subgroup of order d for each divisor d or n. I am not sure what the formal name of this theorem is. I also know Lagrange's theorem (if H a subgroup of G, order of H divides order of G), Sylow's theorem (if G a finite group of order n, then if you take the prime factorization of n, n=p_{1}^{k}p_{2}^{j}...p_{m}^{z} then for each p_{m}^{k} in that factorization G has at least one subgroup of order p_{m}^{i} for 0<=i<=k) I also know another theorem which says if G is finite and Abelian, it has at least one subgroup of order d for every divisor d or n. The thing that gets me is the "at least one subgroup" in these theorems. Are there theorems other than the first one I posted up there which specify exactly how many subgroups of a certain size there are? Like if I have a group of order 500 (or any finite number), say there's no knowledge if it's cyclic or not, is there a way to say exactly how many subgroups it has? What if it's gauranteed to be Abelian? I know if it's Abelian I can say it's isomorphic to direct sums Z_{m} + Z_{n} + ... + Z_{z} for the different combinations of its prime factorization (what I mean by that is say I have an Abelian group of order 24 so its prime factorization is 2*2*2*3, then its isomorphic to Z_{2} + Z_{2} + Z_{2} + Z_{3}, to Z_{4} + Z_{2} + Z_{3}, to Z_{8} + Z_{3}, and to Z_{24}) so do I just then look at the number of subgroups of say Z_{24}? Is there a theorem which would tell me exactly how many subgroups Z_{24} has? 



#2
Nov912, 02:42 PM

P: 350

A google search yielded this paper.
http://math.ubbcluj.ro/~calu/nrsubgroup.pdf It doesn't look simple. 


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