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This following question given in Gallian's algebra text:

What is the smallest positive integer $n$ such that there are two nonisomorphic groups of order $n$?

The answer for this question given in text book is $n=4$ as $\mathbb Z_4$ and $\mathbb Z_2 \times \mathbb Z_2$ served our purpose.I did this by inspection.

Now i wanted to generalise this question i.e; i wanted to know What is the smallest positive integer $n$ such that there are EXACTLY $m$ nonisomorphic groups of order $n$?

Here inspection does'nt work.So please guide me to get to the result.

Thank you!

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  • $\begingroup$ For general $m$, this is an open problem. $\endgroup$
    – James
    Commented May 26, 2017 at 13:53
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    $\begingroup$ Your question is slightly unclear. Do you mean exactly $m$ groups or at least $m$ groups of order $n$. In any case, this question is far too difficult!" See for example math.auckland.ac.nz/~obrien/research/gnu.pdf $\endgroup$
    – Derek Holt
    Commented May 26, 2017 at 13:55
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    $\begingroup$ Fun fact that I hadn't realized, OEIS A000001 is the number of groups of order $n$. $\endgroup$
    – pjs36
    Commented May 26, 2017 at 13:57
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    $\begingroup$ If exactly $m$ was intended, then this function is called ${\rm moa}(n)$ in the referenced paper, although I believe it is still unknown whether there is exists such an $n$ for all $m$. The problem of whether there is an $n$ with ${\rm gnu}(n) = n$ is also open. $\endgroup$
    – Derek Holt
    Commented May 26, 2017 at 14:04
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    $\begingroup$ I like what amounts to be an answer about and to the question above, @DerekHolt I'd love for you to compile them and post an answer. $\endgroup$
    – amWhy
    Commented May 26, 2017 at 14:10

1 Answer 1

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Nobody knows. This is an open problem, at this level of generality. In fact, even the weaker question of whether, given a positive integer $m$, there is an $n$ such that the number of groups of order $n$ is equal to $m$ remains open, as far as I know.

There is a very readable survey on the subject that you might enjoy.

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  • $\begingroup$ Nice answer, @Derek! $\endgroup$
    – amWhy
    Commented May 26, 2017 at 14:07

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