61

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 $n>1$.

Stefan Kohl
  • 19,498
  • 21
  • 73
  • 136
Peter
  • 1,203
  • I asked this question also on math-stack-exchange. – Peter Nov 26 '15 at 23:35
  • 4
    http://math.stackexchange.com/questions/1547948/are-there-n-groups-of-order-n-for-some-n1 – Peter Nov 26 '15 at 23:45
  • 3
    There are no square-free examples, because $N(n) \leq \phi(n)$, for square-free $n > 1$. – James Nov 27 '15 at 00:33
  • 3
    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 $p^{2k^3/27}$). – Noam D. Elkies Nov 27 '15 at 02:08
  • 17
    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.) – Noam D. Elkies Nov 27 '15 at 02:18
  • @Stefan Kohl: Don't give more than 40 upvotes reason to trust in the readabilty of the question ? – Todd Leason Apr 09 '16 at 09:47

1 Answers1

26

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 $72$.

Stefan Kohl
  • 19,498
  • 21
  • 73
  • 136