# Elements in only 1 left/right coset

1. Nov 14, 2006

### Wiemster

My lecture notes state that every element of a group belongs to exactly one right and one left coset of a certain subgroup, but I don't see why this should be the case, so I tries to prove it:

with S and S' elements of the subgroup H of G we then have that if a certain element X of the group G belongs to the coset TH but to T'H as well

i.e. X=TS=T'S' we should have that T=T' right?

This means TS(S'^-1)=T'E with S(S'^-1) again an element of H, but I don't see why T should be T', or did I misintepreted the theorem?

(PS: Somewhat further they use as an example a generelaized torus: Rn / Zn = (R/Z)n which is supposed to b homomorphic to the unit circle. But I don't really see what this 'factor group' represents... )

2. Nov 14, 2006

### ircdan

The right cosets of H in G are the equivalences classes of the equivalence relation ~ on G given by x~y iff xy^-1 in H. They are equivalence classes, so they partition G into a family of mutually disjoint nonempty subsets. So every element in G belongs to exactly 1 right coset(to exactly 1 equivalence class). So if x belongs to two right cosets say Ha and Hb, then it follows that Ha = Hb.

I just said this in words, didn't prove anything. If you want to prove it, you can,

1) Let R be an equivalence relation on a set S. Show every element of S is in exactly one equivalence class.

2) Show the relation defined, ~, is an equivalence relation on a group G(with H being a subgroup).

3) Show the equivalence classes of ~ are the right cosets of H in G.

Then it follows immediately.

3. Nov 15, 2006

### matt grime

Cosets are either equal or disjoint. It is a natural consequence of the definition of a group.

(R/Z) is isomorphic to the unit circle in C. (R/Z)^n is not.