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


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.

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