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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.

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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.

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Note that this also is a counterexample to the claim in the second paragraph. – user641 Apr 2 '11 at 12:49

It's unclear what the original statement was meant to be. Pete's example shows that even if $K$ is abelian, the conclusion does not follow.

The conclusion does follow if $p\lt q$, in which case one does not even require $K$ to be abelian: if $[G\colon K]=p$, then the action of $G$ on the left cosets of $K$ by left multiplication induces a homomorphism $G\to S_p$. Since $p\lt q$, $S_p$ has no elements of order $q$, so any element of order a power of $q$ in $G$ must lie in the kernel of the homomorphism. Since the kernel is contained in $K$ (which is the subgroup of $G$ that fixes the coset $K$), then any $q$-subgroup of $G$, in particular the Sylow $q$-subgroups, is contained in $K$.

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