## 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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