Finite field of with multiplication group isomorphic to the cyclic group of order $17$

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.