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Consider $\mathbb{Z}_n$, where $n$ is any positive integer. Then there is a subgroup of $\mathbb{Z}_n$ for each divisor $d$ of $n$, so that the number of subgroups equals the number of divisors of $n$. If $|subg(\mathbb{Z}_n)| < |div(n)|$, then there's a non-trivial divisor that doesn't generate a proper nontrivial subgroup, etc. So the subgroups of $\mathbb{Z_n}$ correspond to divisors of $n$ in the ring $\mathbb{Z}$. What about other finite groups, is there a ring such that divisors in the ring of the finite group order $|G|$ correspond to subgroups of $G$? Thanks.

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I don't quite understand. Take $A_4$, a group of order 12 that has subgroups of orders $1,2,3,4,12$ but not 6. You want a ring that contains $1,2,3,4,12$ but not 6? I don't think that's possible. – Gerry Myerson Aug 25 '13 at 0:29
A ring such that $6$ doesn't divide $12$, but the others do. – Enjoys Math Aug 25 '13 at 0:30
But $12=6\times2$, and you want 2 to divide 12, so you want 2 to be in the ring, so 6 will divide 12 in the ring. I still don't see a good way to interpret your question. – Gerry Myerson Aug 25 '13 at 0:32
Neither do I, lol :D – Enjoys Math Aug 25 '13 at 0:40
May I suggest that in the future you post only questions that you yourself are able to interpret? – Gerry Myerson Aug 25 '13 at 0:42
up vote 4 down vote accepted

I'm not sure what bearing the ring structure of the integers really has here, unless you want to treat $\mathbb{Z}_n$ as a ring instead of a group in. Cyclic groups are quotients of the additive group of the integers. If we look at a quotient of a group $G$ by a normal subgroup $H$ then normal subgroups in $G/H$ correspond to normal subgroups of $G$ containing $H$. So starting with any group $G$ we can look at the family of quotients of it, and it will have a similar property to the one you described.

The additive group of integers is particularly nice since it is the free group on one generator, so therefore its quotients will be all groups generated by a single element (i.e. cyclic groups). Taking the free group on $n$ generators will have quotients corresponding to all groups generated by $n$ (or fewer) elements. Similarly taking the free abelian group on $n$ generators will have quotients corresponding to all abelian groups on $n$ (or fewer) generators.

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