# Classification of abelianness-forcing numbers

## Contents

## Name

This result is attributed to Dickson, and is hence also called **Dickson's theorem**, though there are many results with that name.

## Statement

The following are equivalent for a natural number :

- Any group of order is an abelian group.
- has prime factorization of the form with for all
**AND**does not divide for any .

## Related facts

## Facts used

## Proof

### (1) implies (2)

It suffices to prove the contrapositive, namely, that violating the conditions of (2) allows one to construct a non-abelian group of that order. There are two subcases.

#### Case of a prime-cube dividing the number

**Given**: A natural number such that there exists a prime number such that divides

**To prove**: There exists a non-abelian group of order

**Proof**: First, by classification of groups of prime-cube order, there exists a non-abelian group of order . Specifically, we can take the unitriangular matrix group:UT(3,p). Call this group .

Now define:

In other words, is the external direct product of this non-abelian group with a cyclic group of order . We obtain that is non-abelian of order .

#### Case of the divisibility condition being violated

**Given**: A natural number , such that there exist distinct primes and a natural number such that divides , and divides .

** To prove**: There is a non-abelian group of order

**Proof**: Note that the product divides .

We can construct a non-abelian group of order as follows: consider the additive group of the field of size . The multiplicative group of this field is a cyclic group of size . Since divides , it has a subgroup of order . Construct a semidirect product of the additive group of order with this subgroup of order .

We can now construct a non-abelian group of order as the external direct product:

### (2) implies (1)

We prove the claim by induction on the number. Note that any divisor of a number that satisfies the condition *also* satisfies the condition.

**Base case for induction** (): Obvious

**Inductive step**: The inductive hypothesis is that the result holds for all smaller numbers.

**Given**: A natural number has prime factorization of the form with for all **AND** does not divide for any . is a group of order .

**To prove**: is abelian

**Proof**:

Step no. | Assertion/construction | Facts used | Give data used | Previous steps used | Explanation |
---|---|---|---|---|---|

1 | Every proper subgroup of is abelian | Fact (1) | inductive hypothesis arithmetic condition on |
Any divisor of satisfies the conditions necessary for the inductive hypothesis to apply. Applying the inductive hypothesis, we obtain that every subgroup is abelian. | |

2 | is metabelian; specifically, either is abelian or it contains an abelian maximal normal subgroup such that the quotient group is cyclic of prime order for some dividing | Fact (2) | Step (1) | Step-fact combination direct | |

3 | If is non-abelian, the maximal normal subgroup has an automorphism of order | Step (2) | [SHOW MORE] | ||

4 | must be abelian | condition on no dividing and all | Step (3) | [SHOW MORE] |