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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, .... , hm} be a subgroup of G.

Let a and b be elements of G and

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

          bH = {bh1, bh2, .... , bhm}

and assume aH = bH.

Since the two cosets aH and bH coincide, each ahi in the coset must correspond to (and be equal to) some bhj in the coset. Thus for each hi ∈H there exists an hj ∈ H such that ahi = bhj in the coset. Now ahi = bhj is equivalent to a-1b = hihj-1 and since hihj-1 ∈H we have proved that a-1b ∈ H.

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

Proof. Assume a-1b = hi ∈H. Then b-1 a = hi-1 ∈H.

Let x = ahj ∈ aH for any hj ∈H. Now


1)        x = ahj = (bb-1)ahj = b(b-1a)hj = b hi-1hj

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

Similarly, bH ⊂ aH since if x = bhk ∈ bH for any hk ∈H, then


2)        x = bhk = (aa-1)bhk = a(a-1b)hk = ahihk

and   ahihk ⊂ 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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