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Prove. A nonempty subset H of a group G is a subgroup of G if and only if a-1b is in H for all a,b in H.

Proof. We prove this by proving two statements:

a) If a-1b is in H for all a,b in H, then H is a group.

b) If H is a group, then a-1b is in H.

Proof of a): To prove a) we assume a-1b is in H for all a,b in H and prove that H meets the four axioms that must hold for a group: 1) closure, 2) associative, 3) existence of identity element, 4) existence of inverse elements. First, since H is a subset of group G, the associative law must hold. Second, the existence of an identity element e in H follows from the fact that since a-1b is in H then necessarily a-1a = e is in H. Third, a-1e = a-1 is in H and every element in H has an inverse. Fourth, for every a,b in H, (a-1)-1b = ab is in H and the Closure Law holds.

Proof of b): Assume that H is a subgroup of G. If a,b is in H, then a-1 is in H and, by the Closure Law, a-1b is in H.