Classification of abelianness-forcing numbers
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 :
- Any group of order is an abelian group.
- has prime factorization of the form with for all AND does not divide for any .
(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
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
|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]|