Prove that the group, under multiplication, of all nonzero elements in a finite field must be a cyclic group.
This is what I did, but I'm not sure if it's right:
First, we look at the additive group G of the field in order to determine the elements in the field. Any finite additive group in a field must be of prime characteristic. Otherwise, if n (where n is not prime) was the characteristic of G, then $n=uv=0 \implies$ either u or v must be zero (since all fields are integral domains), which contridicts the fact that n is the smallest element that equals zero.
If we have a finite additive group of characteristic p in the field, then it must be cyclic (since it is a group of prime order). Now if we want to look at the multiplictative group of the field, we must only look at the invertible elements, which is precisely the group $\Bbb{Z}^{\times}_p \cong \Bbb{Z}_{p-1}$.
Do you think my proof is correct?
Thank you in advance
Best Answer
One way to do this is to look at the group's invariant factor decomposition.
Consider the invariant factor decomposition $\Bbb{F}_{p^n}^{\times} \cong \Bbb{Z}/k_1\Bbb{Z} \oplus \Bbb{Z}/k_2\Bbb{Z} \oplus ... \oplus \Bbb{Z}/k_r\Bbb{Z}$, where $k_{i+1} | k_i$ for all $i$. We wish to show that $\Bbb{F}_{p^n}^{\times}$ is cyclic, or that $r=1$.
The polynomial $x^{k_1} = 1$ must have at most $k_1$ roots. Since, for all $i$, $k_i | k_1$, $x^{k_1} = 1$ for every element $x \in \Bbb{F}_{p^n}^{\times}$. That should be enough for you to find a contradiction if $r>1$.
One possible source of confusion here is that I am writing all groups multiplicatively, using $1$ for the identity, juxtaposition for the group operation and exponents, while the groups $\Bbb{Z}/k\Bbb{Z}$ are usually written additively.