Prove that G cannot have a subgroup H with|H| = n-1, where n= |G|>2. Solution Let G be a group of order n>2 and H a subset of G containing exactly n-1 elements. Then H is obtained by removing exactly one element from G. If this element is the identity, then H clearly cannot be a subgroup, so assume that a non-identity element b is removed. Since n>2, we can choose another non-identity element a (different from b), which must lie in H. By the latin square property, there exists a unique element x in G such that ax = b. Note that x cannot be equal to b, since this would imply that a is the identity, which we assumed it wasn\'t. Thus, x belongs to H as well. If H were a subgroup, then the fact that both a and x are in H would imply that ax = b is in H as well (by closure). However, this is a contradiction. Thus, there is no subgroup of G of order n-1..