Let $G$ be a (probably infinite) group and $H \leq K \leq G$. $|G:H| < \infty \Rightarrow |G:H|=|G:K||K:H|$? Let $G$ be a (probably infinite) group and $H \leq K \leq G$. Is it true that if we have $|G:H| < \infty$ then $|G:H|=|G:K||K:H|$ ?
Thanks in advance.
 A: Recall that $|G:H|$ is the number of left (or right) cosets of $H$ in $G$.  Let $|G:H|=n$, then there are $g_1,\dots,g_n\in G$ s.t. ${g_iH}$ are the distinct cosets of $H$ in $G$.
The first thing to prove is that $|G:K|$ and $|K:H|$ are finite.
First, we show that $|K:H|$ is finite.  Let $K$ be a subgroup of $G$ containing $H$.  Then, since $\cup_i g_iH=G$, some of the $g_iH$'s intersect with $K$.  Suppose $g_1,\dots,g_k$ correspond to the cosets of $H$ which intersect $K$.  Then $g_iH\cap K\not=\emptyset$.  Therefore, for some $h\in H$, $g_ih\in K$.  Let $g_ih=k$.  Since $H$ is a subgroup of $K$, $h^{-1}\in K$, so $g_i=kh^{-1}\in K$.  Therefore, $|K:H|$ is finite and the cosets of $H$ which intersect $K$ are the cosets of $H$ in $K$.
Second, we show that $|G:K|$ is finite.  Consider the set of cosets of $K$ of the form $g_iK$ where $1\leq i\leq n$.  Since $K$ contains $H$, $g_iK\supseteq g_iH$ and since the union of the $g_iH$ is all of $G$, the union of the $g_iK$'s is also all of $G$.  By considering a minimal set of these $g_iK$'s, you have a complete set of cosets.
Finally, since $|G:K|$ is finite, let $g'_1K,\dots,g'_mK$ be the (distinct) cosets of $K$ in $G$, and let $g_1,\cdots,g_k$ be the distinct cosets of $H$ in $K$.  Then the cosets of $H$ in $G$ are $g'_ig_jH$.  You can show that these are distinct and that the number of these is exactly $|G:K||K:H|$.
A: Yes. For reference, you can check out this. I'm assuming $H$ and $K$ are subgroups of $G$.
