Classification of cyclicity-forcing numbers

This article gives a proof/explanation of the equivalence of multiple definitions for the term cyclicity-forcing number
View a complete list of pages giving proofs of equivalence of definitions

The definitions that we have to prove as equivalent

The following are equivalent for a natural number:

There exists exactly one isomorphism class of groups of that order.
Any group of that order is a cyclic group.
Any group of that order is a direct product of cyclic Sylow subgroups.
It is a product of distinct primes $p_{i}$ , such that $p_{i}$ does not divide $p_{j} - 1$ for any prime divisors $p_{i}, p_{j}$ of the order.

Facts used

Finite abelian implies direct product of Sylow subgroups
Finite non-abelian and every proper subgroup is abelian implies not simple: This is the meat of the proof.
Description of automorphism group of cyclic group
Homomorphism between groups of coprime order is trivial
Cyclic over central implies Abelian
Lagrange's theorem
Order of quotient group divides order of group

Proof

Equivalence of definitions (1) and (2)

This follows from two basic observations:

For any natural number $n$ , there exists a cyclic group of order $n$ .
Two cyclic groups of the same order are isomorphic.

Thus, the existence of only one isomorphism class of groups of a given order is equivalent to asserting that every group of that order is cyclic.

Equivalence of definitions (2) and (3)

This follows from the Chinese remainder theorem.

(3) implies (4)

For this, we first prove that the number must be square-free, i.e., it is a product of distinct primes.

Suppose we have a prime factorization as follows:

$n = p_{1}^{k_{1}} p_{2}^{k_{2}} \dots p_{r}^{k_{r}}$ .

Consider the group $G$ that is a direct product of elementary abelian groups of order $p_{i}^{k_{i}}$ . Then, $G$ is an abelian group of order $n$ . Further, if any of the $k_{i}$ is greater than one, then the $p_{i}$ -Sylow subgroup is not cyclic, so $G$ is not cyclic. Thus, if $n$ has a square factor, there is a non-cyclic group of order $n$ . Thus, any cyclicity-forcing number must be square-free.

We thus have:

$n = p_{1} p_{2} \dots p_{r}$

where all the $p_{i}$ are distinct primes.

Now, suppose there exist primes $p_{i}$ and $p_{j}$ such that $p_{i} | p_{j} - 1$ . Then, there exists a non-abelian group $H$ of order $p_{i} p_{j}$ , given as the semidirect product of a cyclic group of order $p_{j}$ , and a cyclic subgroup of order $p_{i}$ in its automorphism group. Let $G$ be the direct product of $H$ with a cyclic group of order $n / (p_{i} p_{j})$ . Then, $G$ is a group of order $n$ . However, $G$ is not cyclic since it has a subgroup isomorphic to $H$ , a non-cyclic group.

Thus, we're forced to have:

$n = p_{1} p_{2} \dots p_{r}$

with $p_{i}$ not dividing $p_{j} - 1$ for any two prime divisors $p_{i}, p_{j}$ of $n$ .

(4) implies (3)

Given: A group $G$ , whose order is $n = p_{1} p_{2} \dots p_{r}$ , where the $p_{i}$ are distinct primes and $p_{i}$ does not divide $p_{j} - 1$ for $i \neq j$ .

To prove: $G$ is cyclic.

Proof: We prove this claim by induction on $n$ . First, note that any divisor of $n$ also satisfies the condition of being square-free as well as the condition that no prime divisor of it divides any other prime divisor minus one.

Base case of induction: The base case of induction, $n = 1$ , is trivial.
Induction hypothesis -- every proper subgroup of $G$ is cyclic: Every proper subgroup of $G$ has order $d$ , for some divisor $d$ of $n$ (by Lagrange's theorem, fact (6)). $d$ is square-free and satisfies the condition that for prime divisors $p, q$ of $d$ , $p$ does not divide $q - 1$ . Further, $d < n$ , so by the inductive hypothesis, every group of order $d$ is cyclic, so the given subgroup is cyclic.
If $G$ is Abelian, then $G$ is cyclic: If $G$ is Abelian, it is a direct product of its Sylow subgroups. Since $n$ is square-free, each of the Sylow subgroups is of prime order, hence cyclic, so $G$ is cyclic.
If $G$ is non-Abelian, then $G$ is not simple: This follows from fact (2).
Any proper normal subgroup of G is central:
1. Suppose $N$ is a proper normal subgroup of $G$ of order $d$ . Then, by fact (2), we know that $N$ is cyclic.
2. Thus, the automorphism group of $N$ has order equal to the Euler-phi function of the order of $N$ . This order is $\prod_{p_{i} | d} (p_{i} - 1)$ .
3. Further, we have a homomorphism from $G$ to $A u t (N)$ given by the action by conjugation. By the assumption on $n$ , we see that the orders of $G$ is relatively prime to the order of $A u t (N)$ , so by fact (4), the homomorphism is trivial. In other words, every element of $G$ acts trivially on $N$ by conjugation.
4. Thus, $N$ is contained in the center of $G$ .
If $G$ is non-Abelian, the center $Z (G)$ of $G$ is nontrivial: This follows by combining steps (4) and (5) above.
If $G$ is non-Abelian, the quotient $G / Z (G)$ is cyclic: The order of the quotient group is a divisor of the order of $G$ (fact (7)), and it is a proper divisor because the center is nontrivial. Hence by the inductive hypothesis, it must be cyclic.
$G$ is Abelian: If $G$ is non-Abelian, then $G / Z (G)$ is cyclic by the previous step. By fact (5), we know that any subgroup that is cyclic over the center of $G$ is Abelian, so we obtain that $G$ itself is Abelian. Thus, we find that even starting with the assumption that $G$ is non-Abelian yields that $G$ is Abelian, so $G$ must always be Abelian.
$G$ is cyclic: This follows by combining steps (3) and (8).

References

Textbook references

Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, Page 149, Exercises 54-55, Section 4.5 (Sylow's theorem), (hints given in exercise)^{More info}