Classification of abelianness-forcing numbers

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


The following are equivalent for a natural number n:

  1. Any group of order n is an abelian group.
  2. n has prime factorization of the form n = p_1^{k_1}p_2^{k_2} \dots p_r^{k_r} with k_i \le 2 for all i AND p_i does not divide p_j^{k_j} - 1 for any 1 \le i,j \le r.

Related facts

Facts used

  1. Lagrange's theorem
  2. Finite non-abelian and every proper subgroup is abelian implies metabelian


(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 n such that there exists a prime number p such that p^3 divides n

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

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

Now define:

G = H \times \mathbb{Z}/(n/p^3)\mathbb{Z}

In other words, G is the external direct product of this non-abelian group with a cyclic group of order n/p^3. We obtain that G is non-abelian of order n.

Case of the divisibility condition being violated

Given: A natural number n, such that there exist distinct primes p,q and a natural number j \in \{1 , 2 \} such that p divides n, and p divides q^j - 1.

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

Proof: Note that the product pq^j divides n.

We can construct a non-abelian group H of order pq^j as follows: consider the additive group of the field of size q^j. The multiplicative group of this field is a cyclic group of size q^j - 1. Since p divides q^j - 1, it has a subgroup of order p. Construct a semidirect product of the additive group of order q^j with this subgroup of order p.

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

G = H \times \mathbb{Z}/(n/pq^j)\mathbb{Z}

(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 (n = 1): Obvious

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

Given: A natural number n has prime factorization of the form n = p_1^{k_1}p_2^{k_2} \dots p_r^{k_r} with k_i \le 2 for all i AND p_i does not divide p_j^{k_j} - 1 for any 1 \le i,j \le r. G is a group of order n.

To prove: G is abelian


Step no. Assertion/construction Facts used Give data used Previous steps used Explanation
1 Every proper subgroup of G is abelian Fact (1) inductive hypothesis
arithmetic condition on n
Any divisor of n satisfies the conditions necessary for the inductive hypothesis to apply. Applying the inductive hypothesis, we obtain that every subgroup is abelian.
2 G is metabelian; specifically, either G is abelian or it contains an abelian maximal normal subgroup N such that the quotient group G/N is cyclic of prime order p for some p dividing n Fact (2) Step (1) Step-fact combination direct
3 If G is non-abelian, the maximal normal subgroup N has an automorphism \sigma of order p Step (2) [SHOW MORE]
4 G must be abelian condition on no p_i dividing p_j^{k_j} - 1 and all k_i \le 2 Step (3) [SHOW MORE]