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

**Point of Post:** In this post we conglomerate and extend a few exercises in Dummit and Foote’s *Abstract Algebra *which will prove that the only positive integers for which the only group (up to isomorphism) of order is are integers of the form are distinct primes with for any .

*Motivation*

This post will complete several lemmas/theorems* *which works towards proving not only that every group of order where for any (greatly generalizing the statement that a group of for primes with is cyclic) but also that numbers of this form are the only numbers for which the converse is true (namely every group of order is cyclic).

*The Classification*

We will need several observations and lemmas to prove this result. The first is a lemma which, in and of itself, is a very powerful theorem:

**Theorem 1: ***Let be a finite group and be such that , then .*

**Proof: **Note first that since is normal each inner automorphism has a well-defined restriction to (i.e. for each the map is a well-defined automorphism). We thus have a map which is easily verified to be a homomorphism. Note then that by definition that where is the centralizer of in (the largest subgroup of for which sits in the center) and so by the first isomorphism theorem we know that embeds into . Thus, we evidently have that divides both and , but since this implies that or . But, by definition this means that .

Another very common, and widely powerful theorem is the following:

**Theorem 2: ***Let be a group and , if (note that containment in the center implies normality) is cyclic then is abelian.*

**Proof: **Let be arbitrary. Since is cyclic we have that and where and . We see then that and for some and so

since were arbitrary the conclusion follows.

We now prove the last ‘common knowledge theorem’ (i.e. something that is widely relevant elsewhere):

**Theorem 3: ***Let be a group of order where are distinct primes. If is abelian then is cyclic.*

**Proof: **This follows trivially from the structure theorem. That said, there is a more simpleminded proof. Namely, by Cauchy’s Theorem there exists with . That said, it’s trivial fact of group theory that if one has a collection of pairwise commuting group elements with pairwise coprime orders then the order of their product is the product of their orders. In particular, since is abelian and for all we have that and so .

Recall now that a simple group is one with no non-trivial proper normal subgroups and maximal subgroups of a group are subgroups which are maximal in terms of containment among proper subgroups (i.e. subgroups such that implies ). Let us make the following observation

**Observation: ***If is a maximal subgroup of a group then or (where denotes the normalizer of in ). In particular, if is not normal in then .*

With this simple observation we can conclude the following

**Theorem 4: ***Let be a finite group and a maximal non-normal subgroup of . Then, the number of non-identity elements of contained in the union over all conjugates of is at most with equality precisely when all the unequal conjugates of have trivial intersection.*

**Proof: **One can quickly check that if and only if or if , or . Thus, the number of distinct conjugates of is equal to , but we know that and so the number of distinct conjugates of is . Moreover, each of these contain precisely non-identity elements and so evidently the cardinality of their union is bounded above by , and clearly equal if and only if all the unequal conjugates of intersect trivially.

With this we can prove the following fascinating theorem:

**Theorem 5: ***Let be a finite group and . Then,*

**Proof: **Find a maximal subgroup of containing (I haven’t proven that this is true, but it follows similarly to the proof of Krull’s theorem). We then have that

and so it suffices to prove that the right hand side of is proper in . To do this we have two choices, either in which case the right hand side of is just which is clearly proper in . If is not normal in then by the previous theorem we have the right hand side is bounded above in cardinality by and so clearly can’t be all of . Either way, we’re done.

**References:**

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

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