Suppose $H$,$K$ are normal subgroups of a group $G$. We don't assume $G$ is finite. Furthermore assume $[G:H]$ and $[G:K]$ are both finite. Then it follows that $[G: H \cap K]$ is divisible by $[G:H]$. Why is this?
Here's what I thought. Suppose we write $G = \bigcup_{i \in I} a_{i}(H \cap K)$ as a disjoint union of left cosets in $H \cap K$, here $I$ is some index set.
Now $a_{i}(H \cap K) \subset a_{i}H$ hence $G \subset \bigcup_{i \in I} a_{i}H$. So we get that $I$ is at least as big as $[G:H]$, thus $[G: H \cap K]$ divides $[G: H]$ no. ? Why is this wrong? why is the other way around?