Abstract Nonsense

Crushing one theorem at a time

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.

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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.

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The Characterizations

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

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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

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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:

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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

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

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

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

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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

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

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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

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\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}

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

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

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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

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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.

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

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

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References:

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

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September 21, 2011 - Posted by | Algebra, Group Theory | , , , , , , , ,

4 Comments »

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