## Unit Group of a Finite Field is Cyclic

**Point of Post:** In this post we prove the basic fact that if is a finite field then is cyclic.

*Motivation*

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

*The Characterizations*

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

**Theorem: ***Let be a finite group, say with , and the property that has at most solutions in for every . Then, is cyclic.*

**Proof: **Let and let . We claim that . Indeed, since for every we have that for every one has that . Thus, clearly there does not exist another otherwise choosing would produce another element of for which which contradicts that there at most solutions to the equation . But, since all Sylow -subgroups are conjugate and we have found that there is only one we may conclude that . What we now claim is that is cyclic. Indeed, by Sylow’s theorems we may find some with . Since then contains solutions to we have that it contains all such solutions. In particular, choosing we find that and since we may conclude from basic theory that and so is cyclic as claimed. Now since all the Sylow -subgroups of are normal we know from basic Sylow theory that where is the unique element of . But since each is cyclic and for we have from basic group theory that is cyclic.

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

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

**Proof: **Let denote the number of cyclic subgroups of of size . We claim now that

where is the Euler totient function. Indeed, evidently

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

That said, note that is equal to which, by basic group theory, is equal to and so

and so our claim laid out in follows by inputting this information into . Now, it is a common fact that

and since we assumed that for each it is easy to deduce from this and that for all . In particular, , but this is equivalent to saying that is cyclic. The conclusion follows.

*Remark: *Although we used it later in the proof, the fact that the sum over of is equal to was not used in the derivation of . In particular, taking so that for all actually implies the more common formula that the sum over of is .

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

**Corollary: ***Let be a finite field, then is cyclic.*

**References:**

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

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