If G is a group of order n, does Lagrange's theorem also imply that there are subgroups of orders dividing n? That is if I have a group of order 15, are there always subgroups of order 1, 3, and 5?
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
The simplest counterexample is that $A_4$, the group of even permutations on 4 items, has 12 elements, but contains no subgroups of order 6.
I believe the general question of when an $mn$-group contains a subgroup of order $n$ is still an open research area. Cauchy's theorem guarantees that if $p$ is prime, then a group of order $pn$ contains an element of order $p$, and therefore a cyclic subgroup of order $p$. So for your example, every group of order 15 does contain subgroups of orders 3 and 5. (And, obviously, 1.) As N.S. noted, the Sylow theorems are important here.
This page discusses the issue in more detail, including some other circumstances when the converse of Lagrange's theorem holds. I think the Wikipedia article on Lagrange's theorem also discusses the converse.
No. It follows from Lagrange Theorem that $G$ has subgroups of order either 3 or 5, but you can't say which....
But in your case, Cauchy Theorem is what you need (or Sylow's theorems).