Does there exist a $n \in \mathbb{N},\; n>1$ such that the number of groups of order $n$ equals $n$?

Me and my friends were just curious and checked the first 1000 numbers (using the internet).

  • 3
    $\begingroup$ See also oeis.org/A000001 $\endgroup$
    – lhf
    Mar 9, 2020 at 16:15
  • $\begingroup$ I check up to order $2047$ and the answer is no. A somewhat close hit above $1000$ is $f(1056)=1028$. $\endgroup$ Mar 9, 2020 at 16:17
  • 2
    $\begingroup$ This is open, and has been discussed many times before here, see for example Is the guess $moa(31)=11774$ in the moa function the true value? $\endgroup$
    – verret
    Mar 9, 2020 at 18:29
  • 4
    $\begingroup$ I don't understand the votes to close. This is a legitimate question (and there isn't much the OP could have tried on their own). $\endgroup$
    – the_fox
    Mar 9, 2020 at 18:44
  • 1
    $\begingroup$ In a paper about gnus, moas and other exotica , the author mentioned the conjecture that there is probably no such $n$ , but the question is , as mentioned , open. The answer is "no" for squarefree numbers , cubefree numbers upto $50\ 000$ , prime powers with exponent not exceeding $4$ and numbers with at most $3$ prime factors (multiplicity considered). This list is not complete, but shows some classes which can be ruled out. $\endgroup$
    – Peter
    May 11, 2020 at 9:59

1 Answer 1


We do not know.

Let $a(n)$ be number of groups of order $n$. There is a conjecture from Conway, Dietrich and O'Brien (see also O'Brien's website here) that the sequence $$n, a(n), a(a(n)), a(a(a(n))), \dots$$ eventually consists of an infinite sequence of $1$.

The number $n$ you are looking for would be a counterexample to this conjecture, which is currently open (and fully checked up to $2047$).

  • 1
    $\begingroup$ To be clear: unless I am not understanding, it's perfectly possible that an elementary argument would show that there was no $n$ of the desired form. I agree that your argument suggests that we will fail to find an example, but a negative solution should still be possible, yes? $\endgroup$
    – lulu
    Mar 9, 2020 at 16:51
  • $\begingroup$ Sure, the non-existence of a fixed point $n$ would not imply that the conjecture is true (you could have cycles or you could have unbounded growth) $\endgroup$ Mar 9, 2020 at 16:55
  • 6
    $\begingroup$ I think this is one of the weirdest conjectures I have ever seen. $\endgroup$ Mar 9, 2020 at 17:34

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