Orders of Quotient Groups (Abstract Algebra)

[lola1990]

Homework Statement

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

Homework 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.

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?

 

[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.

 

[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?

 

[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.

 

[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.

 

[lola1990]

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

 

[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

 

[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.

 

[spamiam]

Firstly, I assume that you mean that S is a set of equivalence classes, and proving $s_{1}=s_{2}$ is proving that $s_{1}\equiv s_{2}$.

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 $a^{-1}b \in K$.

Ok, so if $A(s_{1},t_{1})=A(s_{2},t_{2})$, then $s_{1}t_{1}H=s_{2}t_{2}H$ so that $s_{1}t_{1}\equiv s_{2}t_{2}$. Moreover, since $t_{1}H\subseteq K$ and $t_{2}H\subseteq K$, then $s_{1}\equiv s_{2}$. Indeed, if it were not so then $s_{1}K\cap s_{2}K=\emptyset$since the cosets partition G. But, since $t_{1}H\subseteq K$ and $t_{2}H\subseteq K$, $s_{1}t_{1}H\subseteq s_{1}K$ and $s_{2}t_{2}H\subseteq s_{2}K$ and since $s_{1}t_{1}H=s_{2}t_{2}H$ this is a contradiction. Now, is it kosher to say that since $s_{1}\equiv s_{2}, s_{1}t_{1}\equiv s_{2}t_{2}$ then $t_{1}\equiv t_{2}$?

Most of your proof looks very good. The key fact that you realized was

$s_{1}t_{1}H\subseteq s_{1}K$

but you can actually conclude that $s_1 = s_2$. Can you see why?

Also, I was unsure about proving surjection as well. How do I know that every coset $gH$ can be written as [itex] stH [itex]?

You just have to prove it. You could start by considering gK.

Also, do I hav to prove that this function is well defined? Or is it obvious that if $s_{1}\equiv s_{2}, t_{1}\equiv t_{2}$ then $s_{1}t_{1}\equiv s_{2}t_{2}$

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 ].

 

[spamiam]

Yes, I do understand why $s_{1}=s_{2}$. I just misunderstood what $S$ was. However, can I also conclude from $s_{1}t_{1}H=s_{2}t_{2}H$ that $s_{1}t_{1}=s_{2}t_{2}$? Then I would be home free on the injective portion of the proof.

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.

Just so I make sure I get it, well-defined doesn't have to be proved because there is only one representative from each coset, and our formula is well-defined on Z x Z.

What is Z?

For surjection, consider $gH\in G/H$. Can we just let $s=g$ (since it can come from anywhere in G) and $t=1$? That seems too easy somehow...

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.

oh, also I get the injective proof completely now since $st_{1}H=st_{2}H\Leftrightarrow s^{-1}st_{1}H=s^{-1}st_{2}H\Leftrightarrow t_{1}H=t_{2}H\Leftrightarrow t_{1}=t_{2}$. Is that right?

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

 

[spamiam]

ok, so given gH, consider gK. gK=sK for some s in our set S. Is this the right way to begin? I'm not sure how that helps me.

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

 

[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.

 

[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.

 

[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" ):

However, this bijection is not a natural one, and, in order to define it, we first need to choose a system of coset representatives of H.

"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..."