3
$\begingroup$

Let $p$ be a prime number and $S_p=\{(n!)^2 \bmod p, n=1,2,\dotsc,p-1\}$ the set of residues mod $p$ of squares of factorials. This set is obviously a subset of the group of quadratic residues mod p. For $p=3,5,7,13,17,23,29$ it is also a group for multiplication mod $p$, i.e. a subgroup of the group of quadratic residues.

Question: Are there infinitely many primes $p$ for which $S_p$ is a group modulo $p$?

$\endgroup$
4
  • 2
    $\begingroup$ I found "such primes" a bit difficult to parse, so I re-wrote to what I think you meant. Please feel free to change or revert if I got it wrong. $\endgroup$
    – LSpice
    May 5, 2021 at 17:06
  • $\begingroup$ There are no primes $p$ between 54 and 2000 for which $\# S_p$ divides $p-1$, so my guess would be no. $\endgroup$ May 5, 2021 at 22:37
  • $\begingroup$ @FelipeVoloch, would you define #$S_p$? — sorry for my ignorance. $\endgroup$
    – Wlod AA
    May 7, 2021 at 2:50
  • 2
    $\begingroup$ @WlodAA It's the cardinality of the set in the question. If the set is going to be a subgroup, its order better divide the order of the group. $\endgroup$ May 7, 2021 at 4:11

1 Answer 1

4
$\begingroup$

In view of the identity $(n!)^2/((n-1)!)^2=n^2$, the set $S_p$ generates the subgroup $Q_p<\mathbb F_p^\times$ of quadratic residues; thus, if $S_p$ is a subgroup, then in fact $S_p=Q_p$. Clearly, a necessary and sufficient condition for this to happen is that the sequence of factorials $\{1!,\dotsc,(p-1)!\}$ hits at least one element out of each pair $(a,p-a)$, $1\le a\le (p-1)/2$.

The sequence $\{n!\}_{1\le n\le p-1}$ has been studied by a number of authors (see, for instance, this recent paper and those referenced therein), and is known to be surprisingly difficult. I therefore doubt anything definite can be proved in this direction.

$\endgroup$
5
  • $\begingroup$ The set $ S_p$ satisfy (all operation are mod p) : a. $1 \epsilon S_p$ b. $x\epsilon S_p \to x^{-1}\epsilon S_p$ c. $x,y\epsilon S_p$ and ${x}\neq{y} \to x\cdot{y}\epsilon S_p$ If $S_p$ is a group, we have to proove only that $x\epsilon S_p\to x^{2}\epsilon S_p$ $\endgroup$ May 10, 2021 at 16:42
  • $\begingroup$ @AndrejLeško: a) and b) are quite straightforward, but how do you prove c)? Anyway, if we believe that $S_p$ is "normally" not a subgroup, then (assuming c)) the implication $x\in S_p\ \Rightarrow\ x^2\in S_p$ is just wrong... $\endgroup$
    – Seva
    May 10, 2021 at 19:57
  • $\begingroup$ Correction of my comment: c. $S_p\subseteq S_p\cdot{S_p}$ where $S_p\cdot{S_p}=\lbrace x\cdot{y},x,y\epsilon S_p\rbrace$ $\endgroup$ May 14, 2021 at 16:42
  • $\begingroup$ Well, $S_p\subseteq S_p\cdot S_p$ just because $1\in S_p$, right? $\endgroup$
    – Seva
    May 14, 2021 at 18:17
  • $\begingroup$ Of course,it is trivial, I ment $S_p$ without $1$. It is interesting that all groups $S_p,p<54$ are of the form $\{\pm2^n\}$ or $\{\pm4^n\}$. Why? $\endgroup$ May 15, 2021 at 7:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.