To prove: The number of left cosets of a subgroup is equal to the number of its right cosets.
Proof: Let H be a subgroup of a group G. Let g be an element of G. Consider the map f: H→ gH defined by f(h) = gh for all h in H. We claim that f is a bijection from H to gH.
First, we show that f is injective. Suppose f(h1) = f(h2) for some h1, h2 in H. Then gh1 = gh2, which implies that h1 = h2, by left cancellation law. Therefore, f is injective.
Next, we show that f is surjective. Let gh be an element of gH. Then h is an element of G, since gH is a subset of G. Since H is a subgroup of G, it follows that gh is an element of gH. Therefore, f(h) = gh, which shows that f is surjective.
Hence, f is a bijection from H to gH. Therefore, the number of left cosets of H is equal to the number of right cosets of H.