TITLE: Does the intersection of two finite index subgroups have finite index?
QUESTION [28 upvotes]: Let $(G,*)$ be a group and $H,K$ be two subgroups of $G$ of finite index (the number of left cosets of $H$ and $K$ in $G$). Is the set $H\cap K$ also a subgroup of finite index? I feel like need that $[G\colon(H\cap K)]$ is a divisor of $[G\colon H]\cdot[G\colon K]$, but I dont't know when this holds.
Can somebody help me out?

REPLY [1 votes]: I'm not really adding something to the answers, and surely you already have what you need, but I'll leave this here just for people having the same problem who will eventually end up here.
Since both $H$ and $K$ are finite index subgroups of $G$, we can consider the right cosets $G/H$ and this is finite by hypothesis. In particular $G$ acts (transitively) on $G/H$ by right multiplication, i.e.
$$
 G/H \times G \to G, \qquad
 (Hx, g) \mapsto Hxg.
$$
If we now restrict the action to $K$, and consider the $K$-orbit of $H$, call it $\mathcal{O}_K(H)$, then it must be finite since it's contained in $G/H$. But now by what Isaacs calls the Fundamental Counting Principle (see "Finite Group Theory", theorem 1.4) we exactly know the following equality
$$
 |\mathcal{O}_K(H)| = [K : \text{Stab}_K(H)],
$$
where $\text{Stab}_K(H)$ is the stabiliser of $H$ under the $K$-action. Notice that $\text{Stab}_K(H) = K \cap \text{Stab}_G(H)$, and what's the stabiliser of $H$ for the $G$-action? But of course $H$ itself, so $\text{Stab}_K(H) = K \cap H$, and finally we conclude that $[K : K \cap H]$ is finite. Hence $[G : H \cap K] = [G : K][K : K \cap H]$ must be finite as well.
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