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.