Many months ago, I found a very interesting group theory problem (the *if*-part of the Proposition below) in Lee Zhuo Zhao‘s -signature on Facebook. My first pathetic attempts to solve that problem failed, apparently it required deeper knowledge of group theory than I had at that time. But even after reading some introductory lecture notes on that topic, it took me ages until I succeeded in solving this problem. It wasn’t too difficult then to see that the conclusion is also reversible.

I will prove the following result:

**Proposition.** *Let be a positive integer. Then every finite group of order is cyclic if and only if *

*
**where denotes the Euler’s Totient Function.*

I will use the following notation: If is a group and , then denotes the centraliser of wrt to and denotes the cyclic group generated by .

**Proof of the ***if*-part. Let be a positive integer satisfying .

Note that using the canonical formula for , it immediately follows that is squarefree.

We will use induction on . The claim is trivially true if is a prime number. Suppose now that for all positive integers with , all groups of order are cyclic.

Let be a group of order . We have to prove that is cyclic. From the induction hypothesis and Lagrange’s theorem, it follows that all proper subgroups of are cyclic.

**Lemma 1.** *Any two different elements of a cyclic subgroup of are not conjugate.*

**Proof.** Let be a cyclic subgroup of . Suppose that for integers and some . We claim that .

An immediate induction on shows that

holds for all positive integers . Putting , we see that

,

so divides . Clearly, and divide . Furthermore, is squarefree.

Let be a prime divisor of . Then if and only if . If and , then clearly .

Otherwise, if and are not divisible by , then has a multiplicative inverse modulo , so . Let . Then , so . Also, since by Fermat’s little theorem. But , so . It follows that is a common divisor of and , so . Hence, , so .

Thus, we see that for all prime divisors of and since is squarefree, it follows that . But then , as required.

Next, we prove that the center of is not trivial. Hence, assume for the sake of contradiction that the center of is trivial.

Then for each with , is a proper subgroup of . Furthermore, this subgroup is maximal, for if is a maximal proper subgroup containing (we can assume it to be cyclic by the induction hypothesis), then , so .

It follows that

**Lemma 2.** *For each with , is the unique maximal proper subgroup of containing . *

It is well known that if is a cylic group of order and is a divisor of then contains a unique subgroup of order , namely .

**Lemma 3.** *Let and be two maximal proper subgroups of and let and . Suppose that . Then and are conjugate. In particular, .*

**Proof.** Let be a common prime divisor of and . Then and both contain a unique subgroup and of order respectively. It follows from Lemma 2 that and .

Furthermore, it follows from Sylow’s second theorem that and are conjugate, so for some . Thus, is a generator of , wlog assume that . Now,

.

It follows that , so .

Now, since the center of is trivial, follows from the class equation that

, (1)

where are representatives of the different nontrivial conjugacy classes. Each of the are maximal proper subgroups, the order of each two of them being either equal or coprime (Lemma 3). Furthermore, it is clear that for every prime divisor of , there exists an so that divides . Hence, we can write as , where each of the is the order of some of the . Notice that and for all . Every summand in (1) is of the form for some . Furthermore, for each , each of the elements of belong to different conjugacy classes by Lemma 1. Moreover, for each with , we have that since is a maximal proper subgroup of containing , which is unique by Lemma 2. It follows that in the sum in (1), the summand appears at least times for all . Hence,

so

Thus,

and since for all ,

so , which is the desired contradiction.

It thus follows that the center of is not trivial. Let and . Then and , so by the induction hypothesis, both and are cyclic. Let be a generator of and be a generator of . Let . Then , so . Let . Then .

Assume that . Then divides for some prime divisor of . Since ,

But and is squarefree, so is divisible by exactly one of the numbers and . But then, either

or

which is a contradiction. Hence , so is cyclic, as required.

**Proof of the ***only if*-part. Let be a positive integer satisfying . We have to show that there exists a finite group of order that is not cyclic. The result is obvious if is not squarefree, because if is a multiple prime divisor of , then the group is not cyclic.

Suppose now that is squarefree and . Then there exist prime divisors and of so that is divisible by .

Let be a primitive root modulo . We consider the set

Clearly, since for all if and only if and .

Furthermore, is a subgroup of the symmetric group (the subgroup conditions can readily be verified). Let send to and send to . Then sends to and sends to , so is not abelian.

Consider now the group . Since , . But is not abelian, so is not abelian. In particular, is not cyclic. This completes the proof.

**Remark.** For , the group constructed in the proof of the “only if”-part above is isomorphic to the dihedral group of a regular -gon.

**Further Remark.** I have just noticed that the group is acutally isomorphic to the semidirect product wrt to the homomorphism

,

.