I would like to show that the multiplicative group of a finite field is cyclic. My goal is to use the best of my knowledge to render the argument as efficient yet simple and understandable as possible.

My attempt (edited after reading anon's answer)

Let $\newcommand{\Fq}{\mathbb F_q}K$ be a field. Assume there is an element $\alpha\in K$ of order $d$. Then its powers must generate $d$ distinct elements. In that case we have $$ \langle\alpha\rangle\simeq\newcommand{\Z}{\mathbb Z}\Z/d\Z $$ Furthermore $\alpha$ must be a primitive root of the polynomial $$ X^d-1 $$ Any element of $\langle\alpha\rangle$ raised to the order of the group yields the neutral element. This means that all the $d$ elements of $\langle\alpha\rangle$ are roots of $X^d-1$ so we have found all possible roots.

From this we see that the $\varphi(d)$ generators of $\Z/d\Z$ correspond isomorphically to $\varphi(d)$ generators of $\langle\alpha\rangle$ that must be primitive roots of $X^d-1$. Since $\langle\alpha\rangle$ contains all roots of $X^d-1$ in $K$, these generators must be the only elements of multiplicative order $d$ in $K$.

To sum up so far

A field $K$ contains either NONE or $\varphi(d)$ elements of multiplicative order $d$.

Any finite subgroup of $K^*$ is cyclic

Let $G$ be a subgroup of $K^*$ having $n$ elements. From the cyclic group $\Z/n\Z$ we know that $$ \sum_{d\mid n}\varphi(d)=n $$ Now any element of $G$ must have some order $d$ that divides $n$ by Lagranges theorem. In each such case this provides exactly $\varphi(d)$ elements of order $d$. So if $G$ misses out on any order $d$ dividing $n$ then $G$ will have fewer elements than counted in the sum above which will be a contradiction since $G$ should have $n$ elements.

This shows in particular that $G$ has an element of order $n$ so $G$ is cyclic.

The special case of a finite field

So in a finite field $\Fq$ the multiplicative group $\Fq^*$ is finite thus cyclic by the arguments above.

Any suggestions?

If anyone has suggestions how to simplify this more so that I will not use to much time in my oral exam proving it, I will certainly be grateful!

  • $\begingroup$ Yes, the idea should be that $$\langle\alpha\rangle=\{\alpha,\alpha^2,...,\alpha^d=1\}$$ with multiplication as composition is mapped isomorphically to $\Z/d\Z$ having $+$ as its composition via the isomorphism $$\alpha^k\mapsto k$$ I don't know if this is standard notation, but I suppose it is... $\endgroup$ – String Jan 13 '14 at 10:18

You have two facts at hand:

  1. There are either $0$ or $\varphi(d)$ elements of order $d$ in a field.
  2. Any finite group of units in a field must be cyclic.

In your proof of (1), you assume there is a "primitive root" among the solutions to $X^d-1$, so you are assuming (2). In your proof of (2), you use the fact that there aren't more than $\varphi(d)$ elements of order $d$ in a field, which is part of (1). Ultimately, you are using (1) in your proof of (2) and using (2) in your proof of (1). This is circular reasoning and needs to be fixed.

  • $\begingroup$ Oh, thanks! What if I reverse the first part of the proof of (1) and state that if any $\alpha\in K$ has finite multiplicative order $d$, then $$\langle\alpha\rangle\simeq\Z/d\Z$$ must contain all roots of $$X^d-1$$ and conclude that we have either $0$ or $\varphi(d)$ elements of order $d$ in $K$? $\endgroup$ – String Jan 13 '14 at 10:30
  • $\begingroup$ Then in proof of (2), since $G$ is finite its elements has finite order. This does not assume (1) at all but is a simple fact, right? $\endgroup$ – String Jan 13 '14 at 10:32
  • $\begingroup$ @String: Yes, that combo works and is not circular. $\endgroup$ – anon Jan 13 '14 at 10:39
  • $\begingroup$ Thank you, that straightened out the argument avoiding seemingly circular reasoning. It looks more clear now. $\endgroup$ – String Jan 13 '14 at 10:44

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