Group Theory IV.1 (Cosets)

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

gH := \{gh : h \in H \}.

It turns out that gH and H are disjoint! Indeed, if gh = h’ for some h, h' \in H, 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 g^{-1}g' \in H, then gH = g’H.
  • If g^{-1}g' \not\in H, then gH \cap g'H = \emptyset.

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

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

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