## A Classification of Integers n for Which the Only Groups of Order n are Cyclic (Pt. II)

**Point of Post: **This is a continuation of this post.

With these theorems and observations we can begin the lemmas that will be used to prove the main theorem. Namely:

**Lemma 1: ***Let be a finite simple group in which every proper subgroup is abelian. Prove then that if and are distinct maximal subgroups of then is trivial.*

**Proof:** We begin by noticing that since is abelian that for all and and so normalizes and similarly normalizes . It follows that and but since and are distinct this implies that and so and so by the simplicity of we have that either or is trivial, but clearly the first is impossible (since they are both proper subgroups of ) and so we must conclude the second.

From this we can prove

**Lemma 2: ***Let be a non-abelian group in which every proper subgroup is abelian. Then, is not simple.*

**Proof: **Let , we then have that (otherwise is cyclic, which is not possible since is abelian). We may then find some maximal subgroup of with . By Theorem 5 we have that there exists some with for all . Since we must have that and so there exists some maximal subgroup of with . If either or is normal in we’re done, so assume not. Note first that since for each is maximal (this is clear since conjugation is order preserving both ways) we have by Lemma 1 that if then is trivial, and so by Theorem 4 we have that the number of non-identity elements of contained in some conjugate to is . A similar analysis shows that the number of non-identity elements of contained in some conjugate of is . Now, note that no element of was double counted by counting the number of things in any conjugate of and counting the number of things in any conjugate of . Indeed, this would be to suppose that is non-trivial for some , but by Theorem 4 this would imply that , but this would imply that , and so contradictory to assumption. Thus, we have not double counted as claimed. It follows then that we have counted

distinct elements of (the extra comes from counting the identity element). But, expanding this gives

We claim though that this number is greater than . Indeed, this is equivalent to showing that

or dividing through by gives

but since are non-trivial we have that from where this trivially follows. Thus, we have produced a number of distinct elements of greater than which is clearly impossible.

With these this last lemma we can finally prove our main theorem:

**Theorem: ***Let be the set of all integers such that every group of order is cyclic. Then, if and only if where are distinct primes with for each .*

**Proof: **We first show that each integer of the form described is actually in . To do this we induct on the number of primes in the factorization of such an integer. If then the result follows trivially (since every group of prime order is cyclic). So, suppose that the result is true for each integer whose prime factorization is for distinct primes with and let be distinct primes such that for each and a group of order . By the induction hypothesis we have that every proper subgroup of is cyclic, and so trivially abelian. So, either is abelian and so trivially non-simple or it is non-ablelian from where it follows from Lemma 2 that is non-simple. Either way, we have that there exists some non-trivial proper subgroup . Say that (we may clearly reorder the primes so that this is true) we then have by assumption that is cyclic that (recalling theclassification of the automorphism groups of cyclic groups). But, by assumption this evidently implies that and so by Theorem 1 we know that . We see then that and so cyclic by the induction hypothesis. But, by Theorem 2 this implies that is abelian. But, we may then conclude from Theorem 3 that is cyclic, and so the induction is complete.

Conversely, let not be of this form. We then have two choices, either for some or there exists primes such that . If we can produce non-cyclic groups of order and of order for these two cases then we’ll be done because we can form a group of order by merely taking or cross or respectively which is clearly not cyclic, because if it were then and would have to be cyclic. But, this is easy for since evidently is not cyclic . For the situation is a little harder, but not too bad. Namely, if are distinct primes with we know there exists a non-trivial homomorphism (namely, since we know that there exists some cyclic subgroup with order from where the isomorphism is apparent) and since we know there is by extension a non-trivial homomorphism . Consequently, the semidirect product is a group of order which is not isomorphic to . The conclusion follows.

**References:**

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

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