<P> Let G be a group, let H be a nonempty subset of G and assume that for all a and b in H, ab is in H. To prove that H is a subgroup of G we must show that H is associative, has an identity, has an inverse for every element and is closed under the operation . So, </P> <Ul> <Li> Since the operation of H is the same as the operation of G, the operation is associative since G is a group . </Li> <Li> Since H is not empty there exists an element x in H. If we take a = x and b = x, then ab = xx = e, where e is the identity element . Therefore e is in H . </Li> <Li> Let x be an element in H and we have just shown the identity element, e, is in H. Then let a = e and b = x, it follows that ab = ex = x in H. So the inverse of an element in H is in H . </Li> <Li> Finally, let x and y be elements in H, then since y is in H it follows that y is in H. Hence x (y) = xy is in H and so H is closed under the operation . </Li> </Ul> <Li> Since the operation of H is the same as the operation of G, the operation is associative since G is a group . </Li> <Li> Since H is not empty there exists an element x in H. If we take a = x and b = x, then ab = xx = e, where e is the identity element . Therefore e is in H . </Li>

How to prove h is a subgroup of g