Does there exist infinite group $G$ with subgroups $H,K$ of finite index such that $[G: H \cap K] = [G:H][G:K]$ but $G \ne HK$? Does there exist infinite group $G$ with subgroups $H,K$ of finite index such that $[G: H \cap K] = [G:H][G:K]$ but $G \ne HK$ ? 
 A: Let $U=H\cap K$. Then $U$ is also of finite index in $G$ (and $H$ and $K$). Let $k_1,\ldots k_m\in K$ be representatives of $K/U$, i.e., we have $$K=\biguplus_{i=1}^m k_iU.$$
Since the intersection of finite-index subgroups is of finite index, we conclude that $$N_1:=\bigcap_{i=1}^mk_iUk_i^{-1} $$
is of finite index in $U$ (and the other guys). Moreover $N_1\lhd K$.
Similarly, find $N_2$ of finite index in $U$ with $N_2\lhd H$.
Then $N:=N_1\cap N_2\lhd G$ is of finite index. This allows us to translate the whole question to the finite group $G/N$ and its subgroups.
A: Since $[G:H\cap K]$ and $[G:H]$ exists finitely  , by generalized law of indices of subgroups , we get, 
$[G:H\cap K]=[G:H][H:H\cap K]$ , thus $[H:H\cap K]=[G:K]$ ; now consider 
$\mathcal L_{HK:K} :=\{xK:x \in HK\}$ and $\mathcal L_{H:H\cap K}:=\{y(H \cap K):y\in H\}$ , since 
$x \in H \implies x=xe \in HK$ , so $\mathcal L_{H:H\cap K} \to \mathcal L_{HK:K} ; x(H \cap K) \mapsto xK $ is a well defined map 
( indeed $x(H \cap K)=y(H \cap K) \implies x^{-1}y \in H \cap K \subseteq K \implies xK=yK$) ; now $xK=yK$ , 
where $x,y \in H \implies x^{-1}y \in H , x^{-1}y \in K \implies x^{-1}y \in H \cap K \implies x(H \cap K)=y(H \cap K)$ , 
thus the map is injective , moreover for any $g=hk \in HK  $ , where $h \in H , k \in K$ , 
$h(H \cap K) \mapsto hK=hk(k^{-1}K)=gK $ thus the map is surjective also ; hence $\mathcal L_{HK:K}$ is finite and 
$|\mathcal L_{HK:K}|=|\mathcal L_{H:H\cap K}|=[H:H\cap K]=[G:K]$ . Now since $x \in HK \implies x \in G $ 
so the map $f(xK) = xK  , \forall x \in HK$, from $\mathcal L_{HK:K}$ to the set of all left cosets of $K$ in $G$ is injective 
and 
since we have seen $|\mathcal L_{HK:K}|=[G:K]$ , so this map is surjective also , thus for every $g \in G$ , 
$\exists y=h'k' \in HK$ such that $gK=f(yK)=yK \implies g \in yK  \implies g=yk''=h'(k'k'') \in HK$ , 
thus $G \subseteq HK$ i.e. $G=HK$ 
