Properties of Cosets

Theorem 1: If h \in H, then the right (or left) cosets Hh or hH of H is identical with H, and conversely.

Proof: Let h' be an arbitrary element of H so that hh' \in hH.

Again, since H is a subgroup, we have
h \in H,\,\,\,h' \in H\, \Rightarrow \,hh' \in H

Thus every element of hH is also an element of H.


hH \subset H\, --- \left( i \right)

Again h' = \left( {h{h^{ - 1}}} \right)h' = h\left( {{h^{ - 1}}h'} \right) \in hH

This shows that every element of H is also an element of hH.


H \subset hH\,\, --- \left( {ii} \right)

From (i) and (ii) it follows that hH = H

Similarly, we can show that Hh = H

Conversely, Hh = H \Rightarrow eH \in H \Rightarrow h \in H and similarly hH \Rightarrow H \Rightarrow h \in H.

Theorem 2: Any two right (or left) cosets of H are either disjoint or identical.

Proof: Let H be a subgroup of a group G and let aH and bH be two left cosets. Suppose these cosets are not disjoint. Then they possess an element, say c, in common. Then c may be written as c = ah, and also as c = ah', where h and h' are in H.


ah = bh'

a = bh'{h^{ - 1}}

Since H is a subgroup, h'{h^{ - 1}} \in H.

Let h'{h^{ - 1}} = h'' then a = bh''. Hence aH = \left( {bh''} \right)H \Rightarrow aH = b\left( {h''H} \right) = bH

Therefore the two left cosets are identical if they are not disjoint.

Thus either aH \cap bH = \phi or H = bH

A similar result can be shown to hold for right cosets.

Theorem 3: If H is finite the number of elements in a right (or left) coset of H is equal to order of H.

Proof: The mapping f:H \to Ha, defined by f\left( {{h_i}} \right) = {h_i}a, is obviously onto.

It is one-one, since

f\left( {{h_i}} \right) = f\left( {{h_j}} \right) \Rightarrow {h_i}a = {h_j}a

 \Rightarrow {h_i} = {h_j}

From the right cancellation law; it is onto, since an element ha belonging to Ha is the f - image of h belongs to H.

It follows that the number of elements in a left coset of H is the same as that is H.
Similarly the number of elements inĀ  a left coset of H is the same as that in H.