# Generators of fields, extending groups to fields, finite abelian groups

So I'm working through Koblitz'z "a course in number theory and cryptography" when I came across his proof that every finite field has a generator (ie, There is an element such that the multiplicative group generated by this element are all the units of the field). This implies that the multiplicative group of any finite field is cyclic. So this got me thinking about a number of things. The first being, since the proof only relies on multiplicative commutativity and finiteness, is every finite abelian group cyclic? If not, then there must be a restriction on what finite abelian groups can possibly be the multiplicative group of some finite field. The proof is on page 34 for anyone that has a copy and wants to shed some light on the matter. I'm really just trying to understand what additional structure multiplicative groups of finite fields are endowed with that arbitrary finite abelian groups lack.

Thanks! -wrote using ipad; beware of typos!

• Not every finite abelian group is cyclic, For example, the Klein group $\,\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z\,$ isn't... – DonAntonio Feb 13 '13 at 22:13

The proof must rely on the severe limitations of solutions of polynomials equations in a field: A polynomials of degree $n$ over a field has at most $n$ different solutions in it (in your case probably applied to polynomials of the form $x^d-1$). Arbitrary abelian groups don't pose such limitations on solutions to equations of the form $x^d=e$, which is why the proof fails. Indeed, in the Klein group $\mathbb Z_2\times \mathbb Z_2$, there are four solutions to the equation $x^2=e$, and of course the Klein group is the (smallest) non-cyclic abelian group.
1. Is every finite abelian group cyclic? No! For example, the group $\mathbb{Z}_2^2$ (the Klein four group) is an example of a non-cyclic finite abelian group. The key attribute of the units $U$ of a finite field is that there are at most $d$ elements of order $d$ for each $d\mid |U|$. This is what implies that our group is cyclic!
2. No. Not every finite abelian group can be realized as the units of a finite field. Recall that if $F$ is a finite field then $|F|=p^n$ for some prime $p$. Thus, $|F^\times|=p^n-1$. Thus, a necessary condition is that you are cyclic of order $p^n-1$ for some prime $p$. Moreover, since there is a field of order $p^n$ for every prime $p$ and natural number $n$ this is also sufficient.