Let G be q group containing 21 elements and suppose that H

Let (G, *) be q group containing 21 elements and suppose that H infinity of (G *) such that H notequalto G. Then (H, *) q cyclic subgroup Let (G, *) be a group with 27 elements. Suppose that (G, *) is not to then may not be a cyclic group. You cannot from the given information Ler (6,) = () be a cyclic group of order 6 and let e the 6 G = {a^-6, a^-5, a^-4, a^-3, a^-2, a^-1} Even subgroup is a normal Let (S, *) be a Let H subset of (H, *) is a Let and a = e where identity then a^2 H = a^-6 H group with 84 elements and (H, *) is a normal containing 21 elements that may or may not be an abelian group you cannot form the griven information.

Solution

31) Since order of G is 21 and 21=3*7

So the possible order of H is 3, or 7. Since order of the subgroup divides the order of the group.

As both 3,7 are prime and every group of prime order is cyclic.

Thus H is cyclic