Looking to prove that [G:H][H:K]=[G:K]
Need some support.
In the finite case, this is a fairly trivial application of Lagrange's Theorem. $\displaystyle [G:H]*[H:K]=\frac{|G|}{|H|}*\frac{|H|}{|K|}=\frac{|G|}{|K|}=[G:K]$. Now if G is infinite it is perhaps surprising that this relationship still holds at first, but think about what Lagrange's theorem really tells you and it makes sense. [G:H] is simply the number of partitions of G into cosets (of equal size) of size |H|, then [H:K] is the number of partitions of H into cosets of size |K|. Then think of probability sort of reasoning why it makes sense why when looking to see how many cosets of size |K| we could split G into.
To show two sets have the same cardinality, you can establish a bijection between the two sets. Think of $\displaystyle G/H \oplus H/K$ there is only one natural bijection I can think of, and it works, recall $\displaystyle H \leq G$.
If $\displaystyle G/H$ has $\displaystyle \{ a_i H | i \in I \}$ as the representatives for the cosets and $\displaystyle H/K$ has $\displaystyle \{ b_j K | j \in J \}$ as a representative for the cosets then show that $\displaystyle \{ a_i b_j K| i\in I, j\in J \}$ is a representative for $\displaystyle G/K$. If follows that if $\displaystyle |I| = n$ and $\displaystyle |J|=m$ then $\displaystyle nm$ is the number of left (or right) cosets for $\displaystyle G/K$.
So this is the formal question and my attempt at proving it.
Q:K is a subset of H which is a subset of G, and [G:K] is finite.
Show [G:H] is finite and [G:K]=[G:H][H:K]
(This proof should be valid for G infinite as well).
PF: Since K is a subset of H,
then 1/|K| is greater than 1/|H|.
Therefore |G|/|K| is greater than |G|/|H|.
Since we know that [G:K] is finite, so is [G:H].
Now that we know the index of H in G is finite we can set that equal to a number n.
|G|/|H| = n
Therefore |G|=n|H|
Therefore |G|/|K| = n |H|/|K|
Therefore |G|/n|K|=|H|/|K|
Since [G:K] is finite, we now know that [H:K] is finite.
Therefore |G|/|K| * |H|/|G| = |H|/|K|
Now divide both sides by |H|/|G|.
(Does this proof work in the infinite case?)
I think you are missing the point about the problem with the infinite case. If you have G as an infinite group, say $\displaystyle \mathbb{Z}$ and mod out by a subgroup of finite order, you get something with infinite index. $\displaystyle \mathbb{Z}/<0>\cong \mathbb{Z}$ but what your formula from finite Lagrange no longer makes sense, what is $\displaystyle \frac{|\mathbb{Z}|}{|<0>|}$? $\displaystyle \frac{\infty}{1}$ isn't exactly something you can go around multiplying by other things.
look at these two sets he is saying are the same in cardinality, establish the only bijection here that is obvious $\displaystyle (a_iH,b_jK)\rightarrow(a_i b_jK)$. Now proving what TPH says you need to do amounts to proving this function is both injective and surjective. And it still holds up even if n or m were not finite.
http://feyzioglu.boun.edu.tr/book/chapter2/Ch2(10).pdf this book does the proof on page 104 if you need to see how this is done explicitly.