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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 $g\in H_i$.

Define a map between sets of left cosets byIt is well defined: replacing $g$ by $gu$ for $u\in H\cap K$ changes neither component. It is injective becauseThe Cartesian product on the right has at most $nm$ elements. ThereforeThis proves the basic bound for a finite intersection of finite-index subgroups.