Suppose that $H, K$ are subgroups of a finite group $G$, with $|H|$ relatively prime to $|G:K|$. Does it necessarily follow that $H \leq K$, or is there a counterexample?
This question arose from considering the following. Suppose an abelian subgroup $K$ of $G$ has prime index $p$, and $q \neq p$ is a prime factor of $|K|$. Then any Sylow $q$-subgroup of $G$ is necessarily contained in $K$. I do not know how to show this, but I am guessing that this generalizes to the above statement.
Best Answer
Suppose $G$ is the symmetric group $S_3$ and $K = \{ 1, (12) \}$. Then $[G:K] = 3$. So if what you are asking is true, then $K$ would have to contain all subgroups of $G$ of order $2$. But there are three subgroups of order $2$ and $K$ (obviously!) contains only one of them.