Are infinite indexes multiplicative?

For a group $$G$$ and its subgroup $$H$$, the index of $$H$$ relative to $$G$$, denoted by $$[G:H]$$, is the cardinality of the set $$\{ gH \mid g \in G \}$$.

It is known that $$|G| = [G:H]|H|$$ even if all cardinalities in question are infinite. However, what about the the identity $$[G:H] = [G:K][K:H]$$ for subgroups $$H \subseteq K$$ of $$G$$? Is it true? If so, how to prove it?

• The same proof technique should do. Make a bijection between $G/H$ and $G/K\times K/H$. – Algeboy Feb 4 at 10:31
• @Algeboy Well, constructing a bijection is a standard way to prove an equinumerousity of two sets. The question is, which bijection can be defined between these two sets? :) – Jxt921 Feb 4 at 10:47

Lemma. Let $$X$$ be a set and $$\sim$$ an equivalence relation on it. Assume that for any $$x,y\in X$$ we have a bijection $$f_{x,y}:[x]\to[y]$$. Then for any $$x\in X$$ there exists a bijection $$[x]\times X/\sim$$ $$\to X$$.
$$F:[x]\times X/\sim \; \to X$$ $$F(a, [b])=f_{x,b}(a)$$
Now under the assumption that $$f_{x,y}=f_{x',y'}$$ if $$[x]=[x']$$ and $$[y]=[y']$$, the function is a bijection; I leave that as an exercise. Note that the assumption can be easily enforced by replacing "broken" bijections by a fixed one per pair of equivalence classes. $$\Box$$
Now take $$G/H$$ and define $$\sim$$ on it: $$gH\sim g'H$$ iff $$gK=g'K$$. Note that for any $$g$$ we have a bijection $$K/H\to [gH]$$ given by $$kH\mapsto gkH$$. So our lemma applies, resulting in a bijection between $$G/H$$ and $$K/H\times (G/H)/\sim\,.$$ All that is left is to see that $$(G/H)/\sim$$ is equinumerous with $$G/K$$. Can you complete the proof?