Thread: prove that the cosets Hx and Hy are either equal or disjoint

1. prove that the cosets Hx and Hy are either equal or disjoint

Can anyone verify my proof is ok.

Suppose they have a common element so mx=ky for some m,k E H

This implies that hmx = hky for all h E H

By the latin square property, every element Of H can be expressed as hm or hk for some h.

Therefore Hx =Hy

Cheers

2. I find this post somewhat confusing.

Is H just some subgroup of a group G? Does your problem have anything to do with Latin squares?

If H is just some subgroup of a group G, choose x and y in G, and assume that Hx and Hy are not disjoint. Actually, this seems to be what you have already done, and now you want to prove that the cosets have to be equal.

As they are not disjoint, there is a common element, so mx=ky for some m and k in H, like you said. Then x = m^{-1}ky. Since m^{-1}k is an element of H, we have that x is an element of Hy, whereas any element of Hx is an element of Hy. Indeed, take an element hx of Hx. Since x is in Hy, we have x=dy for some d in H. But then hx = hdy, and as hd is an element of H, we have that hx is an element of Hy.

This proves that Hx is contained in Hy. Exchanging the roles of x and y proves the opposite inclusion.

3. H is a subgroup of G
By the latin square property, I mean every element of H appears once and only once in each row and column of the cayley table. This implies every element of H can be written hm for some h E H. Is there a better word for this?

4. it actually implies more. it implies that every element of H can be written UNIQUELY as hm for a given h in H (we can't have hm = hn, unless m = n), and also

that every element of H can be written uniquely as kh.

the problem with your proof, is that it doesn't involve x and y. here is what you want to do:

suppose Hx ∩ Hy is not empty. then we have hx = h'y, for some h,h' in H.

this means y = h'^-1hx, so y is in Hx. but then for any other element of Hy, say h"y, we have h"y = (h"h'^-1h)x, which is in Hx.

thus Hy is contained in Hx. similarly, since x = h^-1h'y, h"x = (h"h^-1h')y, which is in Hy.

the "latin square property" you refer to only tells you about the behavior of products in H.

unless x and y are IN H, the elements of Hx and Hy will not even be on that table.

5. Originally Posted by Deveno
it actually implies more. it implies that every element of H can be written UNIQUELY as hm for a given h in H (we can't have hm = hn, unless m = n), and also

that every element of H can be written uniquely as kh.

the problem with your proof, is that it doesn't involve x and y. here is what you want to do:

suppose Hx ∩ Hy is not empty. then we have hx = h'y, for some h,h' in H.

this means y = h'^-1hx, so y is in Hx. but then for any other element of Hy, say h"y, we have h"y = (h"h'^-1h)x, which is in Hx.

thus Hy is contained in Hx. similarly, since x = h^-1h'y, h"x = (h"h^-1h')y, which is in Hy.

the "latin square property" you refer to only tells you about the behavior of products in H.

unless x and y are IN H, the elements of Hx and Hy will not even be on that table.
Yeah I can see your proof works, thanks. However I'm talking about the products hm and hk, which are in H. So surely (hm)x=(hk)y means Hx=Hy. I am happy to be proved wrong.

6. i see what you're saying. you're saying we can write an arbitrary element of H,h', as hm, for some h in H, which is true (in fact the element of H which works is h = h'm^-1 which is in H whenever h' and m are, since H is a subgroup). so h'x = (hm)x = h(mx) = h(ky) = (hk)y.

the "latin square property" actually shows this is a bijection between Hx and Hy, and from the above we see that Hx is contained in Hy, so they must be equal.

7. So is my proof is correct since a bijection implies one to one and surjective. I admit your's is clearer