Website owner:  James Miller

Theorem 1. For any two cosets aH and bH of a subgroup H of a group G, aH = bH if and only if a^-1b ∈ H.

Proof. We prove this theorem in two parts:

Part 1. Prove: If aH = bH, then a^-1b ∈ H

Proof. Let H = {h₁, h₂, .... , h_m} be a subgroup of G.

Let a and b be elements of G and

            aH = {ah₁, ah₂, .... , ah_m}

          bH = {bh₁, bh₂, .... , bh_m}

and assume aH = bH.

Since the two cosets aH and bH coincide, each ah_i in the coset must correspond to (and be equal to) some bh_j in the coset. Thus for each h_i ∈H there exists an h_j ∈ H such that ah_i = bh_j in the coset. Now ah_i = bh_j is equivalent to a^-1b = h_ih_j^-1 and since h_ih_j^-1 ∈H we have proved that a^-1b ∈ H.

Part 2. Prove: If a^-1b ∈ H, then aH = bH

Proof. Assume a^-1b = h_i ∈H. Then b^-1 a = h_i^-1 ∈H.

Let x = ah_j ∈ aH for any h_j ∈H. Now

1)        x = ah_j = (bb^-1)ah_j = b(b^-1a)h_j = b h_i^-1h_j

since bb^-1 = e, the identity element. Now bh_i^-1h_j ∈ bH so 1) proves that x ∈ bH and so we have shown that aH ⊂ bH.

Similarly, bH ⊂ aH since if x = bh_k ∈ bH for any h_k ∈H, then

2)        x = bh_k = (aa^-1)bh_k = a(a^-1b)h_k = ah_ih_k

and   ah_ih_k ⊂ aH.

Thus since aH ⊂ bH and bH ⊂ aH, we have proved that aH = bH.

Theorem 2. For any two cosets aH and bH of a subgroup H of a group G, if aH ≠ bH, then aH and bH are disjoint.

Proof. Suppose that aH and bH contain some element c in common. Then c = ah₁ = bh₂ for some h₁ and h₂ ∈ H. This means that h₁h₂^-1 = a^-1b ∈ H and aH = bH by Theorem 1 above.

Thus if two cosets have any element in common, they have all elements in common.

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