# Thread: Coset Proof. Looking for ideas

1. ## Coset Proof. Looking for ideas

The overall problem is this:
If H < N and H, N, and K are subgroups of G, then $HK \cap N = H( K \cap N)$.
I have tried various and sundry methods to work with this, to little result. I have managed to prove that $HK \cap N$ 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: $HK \cap HN = H(K \cap N)$. (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

2. 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.

3. Originally Posted by Deveno
...have fun.
Believe it or not this is the kind of stuff that makes my day.

Thanks, I'll work on it.

-Dan

4. Okay. Now to show that $H( K \cap N ) \subset HK \cap N$.

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

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

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

Thus $hk \in HK \cap N$.

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

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

I think I figured out the best that can be done with my thought that $H ( K \cap N ) = HK \cap HN$. (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 $H( K \cap N ) \subset HK \cap HN$. I'm still looking for the counter-example that proves inequality, though.

Thanks again!

-Dan

5. 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.

6. 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

7. 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.