# Abstract Nonsense

## Unit Group of a Finite Field is Cyclic

Point of Post: In this post we prove the basic fact that if $k$ is a finite field then $U(k)$ is cyclic.

$\text{ }$

Motivation

A very useful fact in both field and group theory is that if $k$ is a finite field then the group of units $U(k)$ is cyclic of order $|k|-1$. There are several ways one may go about doing this. In particular, there are two striking group theoretic facts about $U(k)$ which one hope characterize it. Namely, from basic field theory we know for any polynomial $p(x)\in k[x]$ one has that the number of roots of $p(x)$ in $k$ is bounded above by $\deg(p(x))$. In particular, using only the operation of the group we have that $x^n=1$ has at most solutions in $k$. The other interesting fact, which is easily implied by the bounded amount of roots $x^n=1$, is that $U(k)$ has at most one cyclic subgroup of a given order (this is clearly implied by the previous property since if $|g|=n$ and $C_n$ is the cyclic subgroup generated by $g$ then by Lagrange’s Theorem $x^n=x^{|C_n|}=1$ for each $x\in C_n$ and so, in particular $C_n$ constitutes $n$ solutions of $x^n=1$ and so clearly there can’t be another distinct cyclic subgroup of order $n$). So, we prove that both of these properties characterize cyclic groups and so, in particular, $U(k)$ is cyclic.

$\text{ }$

The Characterizations

$\text{ }$

Our first proof makes clever use of Sylow’s theorems:

$\text{ }$

Theorem: Let $G$ be a finite group, say with $|G|=p_1^{\alpha_1}\cdots p_m^{\alpha_m}$, and the property that $x^n=1$ has at most $n$ solutions in $G$ for every $n\in\mathbb{N}$. Then, $G$ is cyclic.

Proof: Let $k\in[m]$  and let $P\in\text{Syl}_{p_k}(G)$. We claim that $P\unlhd G$. Indeed, since $g^{|P|}=1$ for every $g\in P$ we have that for every $g\notin P$ one has that $g^{|P|}\ne 1$. Thus, clearly there does not exist another $Q\in\text{Syl}_p(G)$ otherwise choosing $g\in Q-P$ would produce another element of $G$ for which $g^{|P|}=g^{|Q|}=1$ which contradicts that there at most $|P|$ solutions to the equation $x^{|P|}=1$. But, since all Sylow $p$-subgroups are conjugate and we have found that there is only one we may conclude that $P\unlhd G$. What we now claim is that $P$ is cyclic. Indeed, by Sylow’s theorems we may find some $H\leqslant P$ with $|H|=p_k^{\alpha_k-1}$. Since then $H$ contains $p_k^{\alpha_k-1}$ solutions to $x^{p_k^{\alpha_k-1}}=1$ we have that it contains all such solutions. In particular, choosing $g\in P-H$ we find that $g^{p_k^{\alpha_k-1}}\ne 1$ and since $g^{p_k^{\alpha_k}}=1$ we may conclude from basic theory that $|g|=p_k^{\alpha_k}$ and so $P$ is cyclic as claimed. Now since all the Sylow $p$-subgroups of $G$ are normal we know from basic Sylow theory that $G\cong P_1\times\cdots\times P_m$ where $P_k$ is the unique element of $\text{Syl}_{p_k}(G)$. But since each $P_k$ is cyclic and $(|P_i|,|P_j|)=1$ for $i\ne j$ we have from basic group theory that $G$ is cyclic. $\blacksquare$

$\text{ }$

Our next theorem shows that the other condition satisfied by $U(k)$ about having at most one cyclic group of each order implies cyclicity. In particular:

$\text{ }$

Theorem: Let $G$ be a finite group and suppose that for each divisor $d$ of $|G|$ one has that $G$ has at most one cyclic subgroup of order $d$ then $G$ is cyclic.

Proof: Let $c_d(G)$ denote the number of cyclic subgroups of $G$ of size $d$. We claim now that

$\text{ }$

$\displaystyle \sum_{d\mid |G|}c_d(G)\varphi(d)=|G|\quad\mathbf{(1)}$

$\text{ }$

where $\varphi$ is the Euler totient function. Indeed, evidently

$\text{ }$

$\displaystyle \sum_{d\mid |G|}\;\;\sum_{x\text{ s.t. }|x|=d}1=|G|\quad\mathbf{(2)}$

$\text{ }$

Now, how do we count how many elements have order $d$ in $G$? Well, note that if $|x|=d$ then $\langle x\rangle$ is a cyclic subgroup of $G$ of order $d$, and moreover this is the only cyclic subgroup of $G$ of order $d$ that contains $x$. Indeed, for if $H$ were another such subgroup then since $x\in H$ we’d have that $G=\langle x\rangle\subseteq H$ and since $|H|=|G|=d$ we’d have that $G=H$. So, the number of elements $x\in G$ with $|x|=d$ is equal to

$\text{ }$

$\displaystyle \sum_{C_d\text{: cyclic subgroup of order }d}\#\{x\in C_d:|x|=d\}$

$\text{ }$

That said, note that $\#\{x\in C_d:|x|=d\}$ is equal to $\#\{x\in C_d:x\text{ generates }C_d\}$ which, by basic group theory, is equal to $\varphi(d)$ and so

$\text{ }$

\displaystyle \begin{aligned}\sum_{C_d\text{: cyclic subgroup of order }d}\#\{x\in C_d:|x|=d\} &=\sum_{C_d\text{: cyclic subgroup of order }d}\varphi(d)\\ & \\ &=c_d(G)\varphi(d)\end{aligned}

$\text{ }$

and so our claim laid out in $\mathbf{(1)}$ follows by inputting this information into $\mathbf{(2)}$. Now, it is a common fact that

$\text{ }$

$\displaystyle \sum_{d\mid |G|}\varphi(d)=|G|$

$\text{ }$

and since we assumed that $c_d(G)\leqslant 1$ for each $d\mid |G|$ it is easy to deduce from this and $\mathbf{(1)}$ that $c_d(G)=1$ for all $d\mid n$. In particular, $c_{|G|}(G)=1$, but this is equivalent to saying that $G$ is cyclic. The conclusion follows. $\blacksquare$

$\text{ }$

Remark: Although we used it later in the proof, the fact that the sum over $d\mid |G|$ of $\varphi(d)$ is equal to $|G|$ was not used in the derivation of $\mathbf{(1)}$. In particular, taking $G=\mathbb{Z}_n$ so that $c_d(G)=1$ for all $d\mid |G|=n$ actually implies the more common formula that the sum over $d\mid n$ of $\varphi(d)$ is $n$.

$\text{ }$

From both of these we get, as justified in the motivation, the following corollary:

$\text{ }$

Corollary: Let $k$ be a finite field, then $U(k)$ is cyclic.

$\text{ }$

$\text{ }$

References:

1.  Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

September 21, 2011 -

1. […] The result has already been proven for and so it suffices to prove this for . To see this note that there is a natural morphism which […]

Pingback by The Unit and Automorphism Group of Z/nZ (Pt. I) « Abstract Nonsense | September 21, 2011 | Reply

2. […] that is cyclic is a little wishy-washy. A more rigorous, and more general proof can be found in this post of […]

Pingback by Groups of Order pq (Pt. II) « Abstract Nonsense | February 28, 2012 | Reply

3. […] just the elementary abelian group . Multiplicatively we know that looks like since, as we have proven before, finite subgroups of the multiplicative group of a field are cyclic. Now, note that this was […]

Pingback by Finite Fields « Abstract Nonsense | April 12, 2012 | Reply

4. […] Proof: If is finite then is finite and the conclusion follows from the cyclicity of . […]

Pingback by Separable Extensions (Pt. II) « Abstract Nonsense | May 4, 2012 | Reply