# Elements of $\mathbb{F}_7^*/\mathbb{F}_7^{*3}$

I think I have forgotten some basic group theory, but I am having hard time representing the elements from $$\mathbb{F}_7^*/\mathbb{F}_7^{*3}$$, where $$\mathbb{F}_7^{*3}$$ denotes all elements that are cubes in $$\mathbb{F}_7^*$$. I have figured out that $$\mathbb{F}_7^{*3} = \{\bar{1},\bar{6}\}$$ and hence $$\mathbb{F}_7^*/\mathbb{F}_7^{*3}$$ is isomorphic to $$\mathbb{Z}/3\mathbb{Z}$$. However, I am looking for representative elements from $$\mathbb{F}_7^*$$. Any help would be appreciated.

• Hint: given two elements of $\mathbb{F}_7^*$, how could you tell if they represent the same element in the quotient group? Remember, this is the same as asking if they lie in the same left coset of $\mathbb{F}_7^{*3}$. – diracdeltafunk Jun 2 at 12:42
• Thanks! I just wrote everything down and it all followed nicely. – Math4Life Jun 2 at 13:09

The group $$\Bbb F_7^\ast$$, as group of units of a finite field, is a cyclic group and it has order 6. It is generated by every primitive root modulo $$7$$, for example, 3. Since $$3^3\equiv 6\pmod 7$$, follows that the quotient group $$\Bbb F_7^\ast/\Bbb F_7^{\ast 3}$$ has representatives $$1,3,9$$.