Classification of abelianness-forcing numbers

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 $n$ :

Any group of order $n$ is an abelian group.
$n$ has prime factorization of the form $n = p_{1}^{k_{1}} p_{2}^{k_{2}} \dots p_{r}^{k_{r}}$ with $k_{i} \leq 2$ for all $i$ AND $p_{i}$ does not divide $p_{j}^{k_{j}} - 1$ for any $1 \leq i, j \leq r$ .

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 $n$ 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:

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

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 $p q^{j}$ divides $n$ .

We can construct a non-abelian group $H$ of order $p q^{j}$ 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:

$G = H \times Z / (n / p q^{j}) Z$

(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 $k_{i} \leq 2$ for all $i$ AND $p_{i}$ does not divide $p_{j}^{k_{j}} - 1$ for any $1 \leq i, j \leq r$ . $G$ is a group of order $n$ .

To prove: $G$ is abelian

Proof:

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 $σ$ of order $p$ .			Step (2)	[SHOW MORE] Pick an element $x$ in $G$ and outside $N$ . Consider the automorphism $σ$ of $N$ induced by conjugation by this element. Since the $p^{t h}$ power of this element is in the abelian subgroup $N$ , the automorphism has order dividing $p_{i}$ . However, since $G$ is non-abelian, the automorphism must be nontrivial (if it were trivial then $N$ and $x$ would centralize each other, and hence generate an abelian group, which would be all of $G$ ).
4	$G$ must be abelian.		condition on no $p_{i}$ dividing $p_{j}^{k_{j}} - 1$ and all $k_{i} \leq 2$	Step (3)	[SHOW MORE] Since $N$ is abelian, all its Sylow subgroups are characteristic, hence $σ$ preserves each of them and its restriction to each has order either 1 or $p$ . If we show that $σ$ restricts to the identity on each Sylow subgroup, then $σ$ would be the identity on all of $N$ , contradicting Step (3). We show this by making two cases on the Sylow subgroups of $N$ . $p = p_{j}$ : In this case, the corresponding Sylow subgroup of $N$ has order $p_{j}^{k_{j} - 1}$ , which is either 1 or $p$ . However, a group of order 1 or $p$ does not have any automorphism of order $p$ , therefore $σ$ must restrict to the identity. $p \neq p_{j}$ : The fixed-point subgroup has order $p_{j}^{l}$ where $0 \leq l \leq k_{j}$ . If $l < k_{j}$ , then the set that is not fixed has size $p_{j}^{l} (p_{j}^{k_{j} - l} - 1)$ . All orbit sizes in this must equal $p$ , so $p$ must divide $p_{j}^{k_{j} - l} - 1$ . If $l < k_{j}$ , then this means $k_{j} - l \in {1, 2}$ . Thus, $p$ divides $p_{j} - 1$ or $p_{j}^{2} - 1$ . Since $p_{j} - 1$ divides $p_{j}^{2} - 1$ , we get that $p$ divides $p_{j}^{2} - 1$ , contradicting our assumption.