# Coset Proof. Looking for ideas

• Apr 10th 2011, 02:04 PM
topsquark
Coset Proof. Looking for ideas
The overall problem is this:
Quote:

If H < N and H, N, and K are subgroups of G, then $\displaystyle HK \cap N = H( K \cap N)$.
I have tried various and sundry methods to work with this, to little result. I have managed to prove that $\displaystyle HK \cap N$ is a subgroup of G, but that was from a tactic that I eventually had to consider unfruitful.

Part of my puzzlement is that the only sign of the existence of G in any of the work I have done is that H, N, and K all have the same identity element. Notable perhaps, but I had expected G to show up more prominently.

I also had the thought that a possible generalization of this theorem exists: $\displaystyle HK \cap HN = H(K \cap N)$. (H is merely a subgroup of G here, not a subgroup of N.) I have been able to construct a number of examples of this, but I admit that my examples were all Abelian. I had hoped that this form might actually be easier to show, but to no avail, even in the Abelian case.

I am looking for ideas, perhaps even at most a general outline of a proof, but not a proof itself. I am having a number of severe difficulties with this section on cosets and I need to work out details on my own for now.

Thanks!

-Dan
• Apr 10th 2011, 02:20 PM
Deveno
those aren't cosets, just subsets of G made from subgroups.

on the LHS, you have elements of the form hk, h in H, k in K. this is also an element of N (because of the intersection sign).

so we have hk = n. but h is in N, since H < N, so h^-1 is in N (N is a subgroup), so k = h^-1n is in N.

thus the k in the hk is in K and N, that is, it is an element of K∩N. this shows HK ∩ N is contained in H(K ∩ N).

the containment the other way is even easier to show. have fun.
• Apr 10th 2011, 02:33 PM
topsquark
Quote:

Originally Posted by Deveno
...have fun.

Believe it or not this is the kind of stuff that makes my day. (Nod)

Thanks, I'll work on it.

-Dan
• Apr 10th 2011, 04:41 PM
topsquark
Okay. Now to show that $\displaystyle H( K \cap N ) \subset HK \cap N$.

Let $\displaystyle k \in K \cap N$. Then we know that $\displaystyle k \in K$ and $\displaystyle k \in N$.

Thus $\displaystyle hk \in H(K \cap N)$.

Now, $\displaystyle hk \in HK$ and $\displaystyle hk \in N$ since $\displaystyle H < N$ and $\displaystyle k \in N$.

Thus $\displaystyle hk \in HK \cap N$.

Finally then $\displaystyle H(K \cap N) \subset HK \cap N$.

Combine this with the work set out in Deveno's post we can say $\displaystyle HK \cap N = H (K \cap N)$.

I think I figured out the best that can be done with my thought that $\displaystyle H ( K \cap N ) = HK \cap HN$. (With H, N, K < G and H is not a subgroup of N.) Using the logic in the above proof I guess the best that can be said is $\displaystyle H( K \cap N ) \subset HK \cap HN$. I'm still looking for the counter-example that proves inequality, though.

Thanks again!

-Dan
• Apr 10th 2011, 05:24 PM
Deveno
let G = D4 = {1,r,r^2,r^3,s,rs,r^2s,r^3s} (r^4 = s^2 = 1, sr = r^3s), let H = {1,r,r^2,r^3}, K = {1,s,r^2,r^2s}, N = {1,rs,r^2,r^3s}.

then K∩N = {1,r^2}, so H(K∩N) = H. but HK = D4, and HN = D4, so HK ∩ HN = D4.
• Apr 10th 2011, 05:58 PM
topsquark
Quote:

Originally Posted by Deveno
let G = D4 = {1,r,r^2,r^3,s,rs,r^2s,r^3s} (r^4 = s^2 = 1, sr = r^3s), let H = {1,r,r^2,r^3}, K = {1,s,r^2,r^2s}, N = {1,rs,r^2,r^3s}.

then K∩N = {1,r^2}, so H(K∩N) = H. but HK = D4, and HN = D4, so HK ∩ HN = D4.

Hmph. Guess I just didn't pick the right groups did I? Well that explains why I couldn't prove it. :)

Thanks again!

-Dan
• Apr 11th 2011, 12:23 PM
Deveno
when looking for counter-examples, try S3 and D4 first. they are small-order, easy to deal with, and (most importantly) non-abelian. D4 has a larger supply of subgroups (good if you need a non-cyclic subgroup).

A4 and A5 make other good candidates for counter-examples. A4 is good for investigating conjugacy classes, and A5 is simple, and provides a handy example when looking for arguments about normality.