# What is the smallest positive integer $n$ such that there are $m$ nonisomorphic groups of order $n$? [closed]

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!

• For general $m$, this is an open problem. Commented May 26, 2017 at 13:53
• 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 Commented May 26, 2017 at 13:55
• Fun fact that I hadn't realized, OEIS A000001 is the number of groups of order $n$. Commented May 26, 2017 at 13:57
• 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. Commented May 26, 2017 at 14:04
• 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. Commented May 26, 2017 at 14:10

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.

• Nice answer, @Derek! Commented May 26, 2017 at 14:07