Cosets
Let H be a subgroup of a group (G, ⋅). We define a relation ρr in the set G as follows:
a ρr b ⇐⇒ ab−1 ∈ H (7.8.1)
Note that if the operation on G were the addition, the relation would be
a ρr b ⇐⇒ a − b ∈ H
which we've already studied in the case G = ℤ, and H = nℤ. For this reason 7.8.1 is known as right congruence modulo H.
The relation 7.8.1 is an equivalence relation on the set G, being reflexive, symmetric and transitive. This equivalence relation divides the set G into disjoint subsets (equivalence classes) known as right cosets modulo H whose union is G.
It is reflexive, since xx−1 = e ∈ H; symmetric, since xy−1 ∈ H implies yx−1 = (xy−1)−1 ∈ H; and transitive, since xy−1 ∈ H, yz−1 ∈ H implies xz−1 = (xy−1)(yz−1) ∈ H.
There are also the left cosets modulo H since we can define the following relation called left congruence modulo H, denoted by ρr:
a ρl b ⇐⇒ b−1a ∈ H (7.8.2)
obtaining the left cosets. In the case G = ℤ, and H = nℤ, the two relations coincide, and so the right and left cosets.
We analyze now how this classes are composed. Let a ∈ G, under the relation ρr we have that
[a] := {b ∈ G | b ρr a} = {b ∈ G | ba−1 ∈ H}
= {b ∈ G | ba−1 = h, h ∈ H}
= {b ∈ G | b = ha, for some h ∈ H} ⊆ Ha.
the converse is easy to prove. Then the class ρr(a) coincides with the subgroup Ha of G, where
Ha := {ha | h ∈ H}
Differently to what we have with conjugacy classes, the following holds true
7.8.1 Proposition. All cosets (left or right) have the same cardinality, which is the cardinality of the subgroup H.
Proof. It suffices to show that there exists a bijective mapping between two arbitrary left cosets (right). This means that all cosets have the same cardinality of H (which is at the same time a right coset and left coset). Let Ha and Hb two arbitrary right cosets. The mapping
ψ: Ha ⟶ Hb
defined as
ψ(ha) = hb
is injective: if ψ(h1a) = ψ(h2a), then it means that h1b = h2b, from which, owing to the cancellation law, h1 = h2;
is surjective: given any element Hb, this will be of the form hb for some h ∈ H, hence having as counterimage ha. □
7.8.2 Example. Let G = 𝓢3 and H the subgroup generated by the transposition (2,3). We calculate right and left cosets, remembering that if H = {a,b,c} then Hx = {ax,bx,cx.}
Right cosets
H = H(2,3) = {id, (2,3)},
H(1,2) = {(1,2), (2,3)(1,2)} = {(1,2),(1,3,2)} = H(1,3,2),
H(1,3) = {(1,3),(2,3)(1,3)} = {(1,3), (1,2,3)} = {(1,3),(1,3,2)} = H(1,2,3)
Left cosets
H = (2,3)H = {id, (2,3)},
(1,2)H = {(1,2), (1,2)(2,3)} = {(1,2),(1,2,3)} = H(1,2,3),
(1,3)H = {(1,3), (1,3)}(2,3)} = {(1,3), (1,3,2)} = {(1,3),(1,3,2)} = H(1,2,3)
Both right and left cosets have the same number of elements (equal to the cardinality of H, i.e. 2). Notice how left and right cosests are not the same
(1,3)H ≠ H(1,3)
(1,2)H ≠ H(1,2)
If we consider the subgroup K = {id, (1,2,3), (1,3,2)}. The right and left cosets with respect to this subgroup are
Right cosets
K = K(1,2,3) = K(1,3,2) = {id, (1,2,3), (1,3,2)},
K(1,2) = {(1,2), (1,2,3)(1,2), (1,3,2)(12)} = {(1,2), (1,3), (2,3)}
= K(1,3) = K(2,3)
Left cosets
K = (1,2,3)K = (1,3,2)K = {id, (1,2,3), (1,3,2)},
(1,2)K = {(1,2), (1,2)(1,2,3), (12)(1,3,2)} = {(1,2), (2,3), (1,3)}
= (2,3)K = (1,3)K
In this case left and right cosets coincide, we shall explain such a behavior later on. ■
The question of when two cosets do coincide is nicely characterized by the following:
7.8.3 Proposition (Equality of cosets). Let G be a group and H be a subgroup of G. If x, y ∈ G, then Hx = Hy if and only if xy−1 ∈ H.
Proof. Suppose Hx = Hy, then since the identity e ∈ H, and since x = ex, the element x ∈ Hx = Hy, so x = hy for some h ∈ H (the equality of sets mean that the two have exactly the same elements). Thus xy−1 = h ∈ H.
Conversely if xy−1 ∈ H, then H (xy−1) = H, hence H (xy−1) y = Hy which leads to Hx = Hy. □