Let’s compare subgroups with subsets; suppose *T* is a subset of *S*, as shown below:

Then an element *s* of *S* is either in *T* or not in *T*. Nothing more could be said. But now, if *H* is a subgroup of* G*, something interesting happens: if *g* is an element of *G* not in *H*, then we define the (left) **coset**

It turns out that *gH* and *H* are disjoint! Indeed, if *gh* = *h’* for some , then *g* = *h’h*^{-1} is also an element of *H*! Thus we get the following picture:

In fact, more generally, any two (left) cosets are either equal or disjoint, as the following theorem shows.

Theorem: letHbe a subgroup ofGandg,g’be any elements ofG.

- If , then
gH=g’H.- If , then .

Since this result is so important, we must provide the proof.

- Suppose . Any element of
*g’H*can be written as*g’h*, for some ; but then*g’h*=*g*(*g*^{-1}*g’*)*h*and (*g*^{-1}*g’*)*h*is an element of*H*by assumption. So and we have . The reverse inclusion follows from*gh*=*g’*(*g’*^{-1}*g*)*h*and (*g’*^{-1}*g*) = (*g*^{-1}*g’*)^{-1}. - Suppose . Any element in the intersection can be written as
*gh*=*g’h’*for some . But this means*g*^{-1}*g’*=*hh’*^{-1}which lies in*H*, thus contradicting our assumption. Hence, is empty.

Finally, every element *g* of the group *G* must occur in some (left) coset, namely *gH*. This establishes the main point of the section:

The left cosets partition

Ginto disjoint subsetsgH.

Pictorially, we have: