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. . 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 there is only one natural bijection I can think of, and it works, recall .
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|/|K| = n |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 and mod out by a subgroup of finite order, you get something with infinite index. but what your formula from finite Lagrange no longer makes sense, what is ? isn't exactly something you can go around multiplying by other things.
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.