61
$\begingroup$

Given a positive integer $n$, let $N(n)$ denote the number of groups of order $n$, up to isomorphism.

Question: Does $N(n)=n$ hold for some $n>1$?

I checked the OEIS-sequence https://oeis.org/A000001 as well as the squarefree numbers in the range $[2,10^6]$ and found no example. Since we have many $n$ with $N(n)<n$ and some $n$ with $N(n) \gg n$, I see no reason why $N(n)=n$ should be impossible for $n>1$.

$\endgroup$
6
  • $\begingroup$ I asked this question also on math-stack-exchange. $\endgroup$
    – Peter
    Commented Nov 26, 2015 at 23:35
  • 4
    $\begingroup$ math.stackexchange.com/questions/1547948/… $\endgroup$
    – Peter
    Commented Nov 26, 2015 at 23:45
  • 3
    $\begingroup$ There are no square-free examples, because $N(n) \leq \phi(n)$, for square-free $n > 1$. $\endgroup$
    – James
    Commented Nov 27, 2015 at 0:33
  • 3
    $\begingroup$ Note, though, that $N(2^k)$ grows much faster than $2^k$ (something like $2^{2k^3/27}$ if I remember right). Likewise for $N(p^k)$ for any fixed prime $p$ (when $N(p^k)$ grows like $p^{2k^3/27}$). $\endgroup$ Commented Nov 27, 2015 at 2:08
  • 17
    $\begingroup$ Also: wow, I didn't know (or didn't remember) that this was the very first OEIS sequence; what I expected would be the first is either #12 or #27. (Question edited only to fix spelling and change ">>" to "\gg" in the TeX.) $\endgroup$ Commented Nov 27, 2015 at 2:18

1 Answer 1

27
$\begingroup$

A "near-miss" is $N(19328) = 19324$, while the only $n \leq 2000$ such that $|N(n)-n| \leq 25$ are $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $10$, $11$, $12$, $13$, $14$, $15$, $16$, $17$, $18$, $19$, $20$, $21$, $22$, $23$, $24$, $25$, $26$, $27$, $28$, $32$, $36$, $48$, and $72$.

$\endgroup$

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.