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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^{-1}b ∈ H.

Proof. We prove this theorem in two parts:

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

Proof. Let H = {h_{1}, h_{2}, .... , h_{m}} be a subgroup of G.

Let a and b be elements of G and

aH = {ah_{1}, ah_{2}, .... , ah_{m}}

bH = {bh_{1}, bh_{2}, .... , 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^{-1}b = h_{i}h_{j}^{-1} and since h_{i}h_{j}^{-1} ∈H we have proved that a^{-1}b ∈ H.

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

Proof. Assume a^{-1}b = 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^{-1}a)h_{j} = b h_{i}^{-1}h_{j}

since bb^{-1} = e, the identity element. Now bh_{i}^{-1}h_{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^{-1}b)h_{k} = ah_{i}h_{k}

and ah_{i}h_{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 = ah_{1} = bh_{2} for
some h_{1} and h_{2} ∈ H. This means that h_{1}h_{2}^{-1} = a^{-1}b ∈ 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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