Show that `G={1,w,w^2}` where `w` is a cube root of unity is a cyclic group.

It suffices to show that the only subgroup `H` of `G` that contains `w` is `G` itself.

Let `H<=G,winH` . Now the identity element of the group is the identity element for every subgroup, so `1inH` .

H is a subgroup if and only if it is closed under multiplication. If H={1,w} then it is not closed, since w^2 is not in H. Thus H must be {1,w,w^2} which is G.

Then G is cyclic.

-------------------------------------------------------------------

Another approach is to show that G is a group that is generated by w. w^0=1,w^1=w, w^2=w^2,w^3=1

-------------------------------------------------------------------

Another approach would be to show that G is isomorphic to Z_3 , the group formed by {0,1,2} with operation addition modulo 3.