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 : let H be a subgroup of G and g, g’ be any elements of G.
- 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-1g’)h and (g-1g’)h is an element of H by assumption. So and we have . The reverse inclusion follows from gh = g’(g’-1g)h and (g’-1g) = (g-1g’)-1.
- Suppose . Any element in the intersection can be written as gh = g’h’ for some . But this means g-1g’ = 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 G into disjoint subsets gH.
Pictorially, we have: