# 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:

1. There exists exactly one isomorphism class of groups of that order.
2. Any group of that order is a cyclic group.
3. Any group of that order is a direct product of cyclic Sylow subgroups.
4. 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.
5. It is relatively prime to the value of its Euler totient function.

## Facts used

1. Finite abelian implies direct product of Sylow subgroups
2. Finite non-abelian and every proper subgroup is abelian implies not simple: This is the meat of the proof.
3. Description of automorphism group of cyclic group
4. Homomorphism between groups of coprime order is trivial
5. Cyclic over central implies abelian
6. Lagrange's theorem
7. 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: $n = p_1p_2 \dots p_r$

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 $p_ip_j$, 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 $n/(p_ip_j)$. 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: $n = p_1p_2 \dots p_r$

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 $n = p_1p_2 \dots p_r$, where the $p_i$ are distinct primes and $p_i$ does not divide $p_j - 1$ for $i \ne j$.

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.

Proof of inductive step:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Every proper subgroup of $G$ is cyclic Fact (6) inductive hypothesis, arithmetic condition on $n$ [SHOW MORE]
2 If $G$ is non-abelian, $G$ is not simple Fact (2) Step (1) [SHOW MORE]
3 Any proper normal subgroup of $G$ is central Fact (4) For any two prime divisors $p_i, p_j$ of $n$, $p_i$ does not divide $p_j - 1$ Step (1) [SHOW MORE]
4 If $G$ is non-abelian, the center $Z(G)$ of $G$ is nontrivial Steps (2), (3) [SHOW MORE]
5 If $G$ is non-abelian, the quotient $G/Z(G)$ is cyclic Fact (7) inductive hypothesis Step (4) [SHOW MORE]
6 $G$ is abelian Fact (5) Step (5) [SHOW MORE]
7 $G$ is cyclic $n$ is square-free Step (6) [SHOW MORE]