# Orders of Quotient Groups (Abstract Algebra)

1. Oct 21, 2011

### lola1990

1. The problem statement, all variables and given/known data
Let H be a subgroup of K and K be a subgroup of G. Prove that |G|=|G:K||K|. Do not assume that G is finite

2. Relevant equations
|G|=|G/H|, the order of the quotient group of H in G. This is the number of left cosets of H in G.

3. The attempt at a solution
I would use LaGrange's Thm, but G is not necessarily finite. I thought a good idea would be to try to find an isomorphism from G/K x K/H to G/H. I defined A(aK x bH)=abH, but I am having trouble proving it is an isomorphism (not even sure it is one!). Is this the right approach? If not, what would be a better way?

2. Oct 22, 2011

### spamiam

I remember this one being tricky. Your approach seems good. I think I had to use a set of coset representatives rather than the cosets themselves when defining the bijection.

3. Oct 22, 2011

### lola1990

How would that bijection be defined? So if [a] is the set of representatives for aK and is the set of representatives for bH, where all elements of [a] are in G and all elements of are in K, we want to get to some [c] where [c] is in G and is the set of representatives for cH. How would you do that?

4. Oct 22, 2011

### spamiam

I'm suggesting this because I ran into the same thing when proving this statement. With your map defined as in the first post, try to show it's one-to-one. If I recall correctly, you'll run into trouble and hopefully see where you can use coset representatives to solve this problem. I'm being vague for the moment--post your attempt at proving your map is a bijection and I'll be able to give more hints.

5. Oct 22, 2011

### lola1990

ok, I'll give it a shot. Suppose that A(aK x bH)\neq A(xK x yH), so that abH\neq xyH. Thus, (ab)^{-1}(xy)\notin H. Then, a^{-1}b^{-1}xy\notin H. This is where I am stuck... I want to show that either a^{-1}x\notin K or b^{-1}y\notin H but I'm not sure how to do that.

6. Oct 22, 2011

### lola1990

oh, sorry I meant b^{-1}a^{-1}xy\notin H

7. Oct 22, 2011

### lola1990

You could also say that ab\notin xyH and xy\notin abH. Then, we want to show that a\notin xK or b\notin yH. This is equivalent to showing that a\in xK implies b\notin yH or that b\in yH implies a\notin xK. Assume that b\in yH and a\in xK... now I'm not sure where to go

8. Oct 23, 2011

### spamiam

Here, try defining your map via coset representatives as I suggested before. Let S be a set of left coset reps for K in G and T be a set a of left coset reps for H in K, and define $A : S \times T \to G/H$ by $A(s,t) = stH$. Also, I think equality is easier to prove than inequality, so try assuming $A(s_1, t_1) = A(s_2,t_2)$ and show that $s_1=s_2$ and $t_1=t_2$. And remember that $H \subseteq K$.

N.B. If you put [ itex ] and [ /itex ] (without the spaces) around your TeX, it will produce nice LaTeX, e.g. $b \notin yK$. You can right-click on my example to see the markup.

9. Oct 23, 2011

### spamiam

A coset is an equivalence class. What I meant is this: each coset has at least one element, perhaps many elements. Pick one for each coset of K in G (using the axiom of choice, if necessary) and call this collection S. Do the same for each coset of H in K and call this collection T. The reason this helps is that there is for each coset aK of K in G, there is a unique element of S which is an element of aK, i.e. a representative of aK. Thus if we have aK = bK and $a,b \in S$, then we actually know that a=b, rather than just knowing $a^{-1}b \in K$.

Most of your proof looks very good. The key fact that you realized was
but you can actually conclude that $s_1 = s_2$. Can you see why?

You just have to prove it. You could start by considering gK.
No, you don't have to prove it's well-defined. Since there is only one representative from each coset, this isn't a problem unlike when working with cosets themselves.

You just forgot a [ /itex ] somewhere, so your text appeared as it would in a math environment. In fact I can see that after writing stH, you forgot the slash in the [ /itex ].

10. Oct 23, 2011

### spamiam

You don't know that $s_1 t_1$ is in either S or T, so no, you have to rely on your previous proof.

What is Z?

Who says g is in S and 1 is in T? Like I said, given a coset gH of H in G, start by considering gK.

Looks good. (I corrected the TeX a bit.)

11. Oct 23, 2011

### spamiam

That's how I began. What does this tell you about g?

12. Oct 23, 2011

### spamiam

All right, one more hint before I go to bed. You're on the right track. So far in the onto proof, you've used the definition of S. How can you use T in a similar way? Remember, you're trying to find s and t such that gH = stH.

13. Oct 23, 2011

### lola1990

got it. g=sk for some k in K, and k is in tH for some t. Thus, k=th for some h and g=sth so that g is in stH and gH=stH.

14. Oct 24, 2011

### spamiam

I remember that I had a really tough time doing this proof because I kept trying to define my maps using the cosets themselves, rather than using a set of representatives. Apparently there is some deeper reason for this (from http://groupprops.subwiki.org/wiki/Index_of_a_subgroup" [Broken]):
"Natural" has a precise meaning in category theory (which I know very little about), but I've heard that one can think of something natural as "coordinate-free." I guess this isn't that unusual though, considering how many proofs in linear algebra start with, "Choose a basis $\beta$ for the vector space V..."

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