# Finite field of 17 elements cyclic under multiplication

I am asked whether there exists a finite field whose non-zero elements form a cyclic group of order 17 under multiplication.

I can't seem to get comfortable with the idea of finite fields, and I know that the field is unique and I think it is isomorphic to $\mathbb{Z}/17\mathbb{Z}$, however am not sure how to proceed. Is there something I can extract from the cyclic group structure under addition of $\mathbb{Z}_{17}$ to its multiplicative group structure?

• But $0$ wouldn't be an element of the group, so that would only give you 16 elements. You'd need a finite field of order 18. – saulspatz Apr 23 '18 at 19:16

## 3 Answers

The nonzero elements of a finite field always form a cyclic group under multiplication. Finite fields always have order $p^n$ for some prime $p$ and positive integer $n$.

Combine these two facts to see this is not possible.

• Thanks, I get how to do this problem from your answer. Do the nonzero elts form a cyclic group under multiplication for group order $p^n$ because they are elements of the splitting fields of $x^{p^n} - x$ which are just successive powers of the $p^nth$ root of unity? – blanchey Apr 23 '18 at 19:19
• no problem. Yes, but really they are the roots of $x^{p^n-1}-1$. – qbert Apr 23 '18 at 19:24

Proof that the multiplicative group $\boldsymbol{F^\times}$ of a finite field $\boldsymbol F$ is cyclic:

Remember that in a group, the exponent is the least positive integer $n$, if it exists, such that $x^n = 1$ (the unit element of the group). This of course implies all elements in the group have finite order. It is equivalent to say all elements have finite order and the orders have a least common multiple (which is not necessarily true if the group is infinite).

In a finite group, with exponent $e$, there exists an element $\alpha$ of order $e$. So in a finite field, the non-zero elements satisfy the equation $x^e-1=0$ If $F$ is a field with order $q$ ($=p^n$), the equation $x^e-1=0$ has at most $e$ solutions, whence $e\ge q-1$ and finally $e=q-1$, so $F^\times$ is generated by $\alpha$.

Note:

One can actually prove more:

In a field $F$ (finite or infinite), any finite subgroup of the multiplicative group $F^\times$ is cyclic.

• Yes, very good to have this here. Perhaps add the definition of exponent for those who forgot? As I recall it is the lowest common multiple of the orders of the elements of the finite group. – Vincent Apr 23 '18 at 19:59
• @Vincent: You're right. I've added a few words. – Bernard Apr 23 '18 at 20:12

A finite field whose non-zero elements form a group of order $17$ would have $18$ elements. However, the number of elements in a finite field is always a power of a prime, which $18$ is not.