Prove that the set {1,w,w^2}, where w(omega) is a cube root of unity, is a cyclic group with respect of multiplication. 

1 Answer | Add Yours

embizze's profile pic

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

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.

We’ve answered 317,895 questions. We can answer yours, too.

Ask a question