Abelianness-forcing number Q

6

In the theory of finite abstract group, abelianness-forcing number $n$ is characterized as a positive integer with standard factorization $n=p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}$ with $k_i \le 2$ and $p_i$ does not divide $p_j^{k_j}-1$ for any $1 \le i,j \le r$. I want to define a function "AbeliannessForcingNumberQ" which returns "True" if and only if the argument is a abeliannes-forcing number. But I cannot figure out how to deal with prime factors and exponents in a given number. Please help to define this function.

seoneo

Posted 2019-07-09T22:45:45.137

Reputation: 163

The following is an external link for the corresponding mathematical concept. https://groupprops.subwiki.org/wiki/Abelianness-forcing_number

– seoneo – 2019-07-09T22:46:39.390

Answers

6

abeliannessForcingNumberQ[n_Integer] := And @@ Flatten[{
   Table[Not@Divisible[Power @@ p - 1, q], {p, FactorInteger[n]}, {q, First /@ FactorInteger[n]}], 
   Thread[Last /@ FactorInteger[n] <= 2]
   }]

Select[Range[20], abeliannessForcingNumberQ]
(* 2, 3, 4, 5, 7, 9, 11, 13, 15, 17, 19 *)

See also A051532.

AccidentalFourierTransform

Posted 2019-07-09T22:45:45.137

Reputation: 9 042

2What a quick and exact response! – seoneo – 2019-07-09T23:12:32.907

As a minor correction, we should include $1$ as an abelianness forcing number. – seoneo – 2019-10-28T04:24:02.043