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 $n$:

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 $k_{i}\leq 2$ for all $i$ AND $p_{i}$ does not divide $p_{j}^{k_{j}}-1$ for any $1\leq i,j\leq r$.

Related facts

• Classification of cyclicity-forcing numbers
• Classification of nilpotency-forcing numbers

Facts used

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

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:

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

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 $k_{i}\leq 2$ 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 $\sigma$ of order $p$			Step (2)	[SHOW MORE] Pick an element $x$ in $G$ and outside $N$. Consider the automorphism $\sigma$ of $N$ induced by conjugation by this element. Since the $p^{th}$ 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 $\sigma$ preserves each of them and its restriction to each has order either 1 or $p$. If we show that $\sigma$ restricts to the identity on each Sylow subgroup, then $\sigma$ 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 $\sigma$ 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.