1. need help with proof

Looking to prove that [G:H][H:K]=[G:K]

Need some support.

2. Lagrange

In the finite case, this is a fairly trivial application of Lagrange's Theorem. $[G:H]*[H:K]=\frac{|G|}{|H|}*\frac{|H|}{|K|}=\frac{|G|}{|K|}=[G:K]$. 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.

3. i completely understand the first argument if G is finite. It is obvious that the indexes can be represented as fractions as you neatly did.

But for the infinite case, what else is expected to be shown for a strong proof?

4. Cardinality

To show two sets have the same cardinality, you can establish a bijection between the two sets. Think of $G/H \oplus H/K$ there is only one natural bijection I can think of, and it works, recall $H \leq G$.

5. If $G/H$ has $\{ a_i H | i \in I \}$ as the representatives for the cosets and $H/K$ has $\{ b_j K | j \in J \}$ as a representative for the cosets then show that $\{ a_i b_j K| i\in I, j\in J \}$ is a representative for $G/K$. If follows that if $|I| = n$ and $|J|=m$ then $nm$ is the number of left (or right) cosets for $G/K$.

6. Infinity

What if one or both subgroups were subgroups of infinite index? is it really safe to just say that without actually establishing the bijection? I mean it is a formality I guess, but that was the bijection I was thinking of.

7. Originally Posted by Gamma
To show two sets have the same cardinality, you can establish a bijection between the two sets. Think of $G/H \oplus H/K$ there is only one natural bijection I can think of, and it works, recall $H \leq G$.
Originally Posted by Gamma
What if one or both subgroups were subgroups of infinite index? is it really safe to just say that without actually establishing the bijection? I mean it is a formality I guess, but that was the bijection I was thinking of.
But you are assuming that $K$ is a normal subgroup of $H$ and $H$ is a normal subgroup of $G$.
That is why you are able to construct a bijection.

8. Touche

Yeah, you are right, to get that direct sum they need to be normal. I guess I didn't even think about quotienting out by non normal subgroups, well put.

9. 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?)

10. problem

I think you are missing the point about the problem with the infinite case. If you have G as an infinite group, say $\mathbb{Z}$ and mod out by a subgroup of finite order, you get something with infinite index. $\mathbb{Z}/<0>\cong \mathbb{Z}$ but what your formula from finite Lagrange no longer makes sense, what is $\frac{|\mathbb{Z}|}{|<0>|}$? $\frac{\infty}{1}$ isn't exactly something you can go around multiplying by other things.

Originally Posted by ThePerfectHacker
If $G/H$ has $\{ a_i H | i \in I \}$ as the representatives for the cosets and $H/K$ has $\{ b_j K | j \in J \}$ as a representative for the cosets then show that $\{ a_i b_j K| i\in I, j\in J \}$ is a representative for $G/K$. If follows that if $|I| = n$ and $|J|=m$ then $nm$ is the number of left (or right) cosets for $G/K$.
look at these two sets he is saying are the same in cardinality, establish the only bijection here that is obvious $(a_iH,b_jK)\rightarrow(a_i b_jK)$. 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.