Finite-index subgroup

A subgroup $H\le G$ has finite index when the coset space $G/H$ is finite. Intersecting a finite-index subgroup with a subgroup again gives a finite-index subgroup of the latter.

In an infinite group $G$, the complement of a finite union $H_1\cup \cdots \cup H_k$ of finite-index subgroups is either empty or infinite. To prove this, their intersection $L$ has finite subgroup index by the finite intersection of finite-index subgroups bound, so it is infinite: otherwise finitely many finite left cosets would cover $G$. If $g$ lies outside the union, the whole infinite left coset $gL$ does too. Indeed, $g\ell \in H_i$ and $\ell \in L\subseteq H_i$ would imply $g\in H_i$.

For finite-index subgroups $H,K$ of a group $G$, the map of sets of left cosetsis well defined and injective: equality of both components is equivalent to the difference $g^{\prime -1}g$ belonging to $H\cap K$. Hence the displayed subgroup index bound holds, without a normal subgroup hypothesis. By mathematical induction, the index of a subgroup for a finite intersection is at most the product of the individual subgroup indices.