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This question already has an answer here:

Suppose $n=18$, then all possible groups of order $18$ is $5$. Among them $2$ are abelian and $3$ are non-abelian.

Let $n$ be a natural number. How can I determine the number of all possible groups (abelian and non-abelian) of order $n$?

Is there any theorem or result that can determine number of all possible groups of given order $n$?

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marked as duplicate by Zev Chonoles, anon, joriki, Davide Giraudo, Derek Holt group-theory Jul 29 '15 at 9:17

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    $\begingroup$ You can't, no. There is no closed formula for the number of groups up to isomorphism which have given order $n$, except for small $n$, or very special cases. $\endgroup$ – the_fox Jul 29 '15 at 4:06
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    $\begingroup$ There's an OEIS entry (it's number 1!) for this, certainly no closed-formula. You'll notice occasional large spikes; i.e., 267 groups of a certain order, in that entry. That order is $64$: the number of groups of order $2^n$ grows extremely quickly (as in this OEIS entry. $\endgroup$ – pjs36 Jul 29 '15 at 4:06
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    $\begingroup$ @pjs36 I remember reading somewhere more than 99% groups of order $<2000$ are of order $1024$. $\endgroup$ – Eoin Jul 29 '15 at 4:19
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    $\begingroup$ However, you can determine the number of abelian groups of a given order (by the fundamental theorem). Non-abelian groups are much harder though, and the analysis usually begins with Sylow theory. $\endgroup$ – Prahlad Vaidyanathan Jul 29 '15 at 4:25
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    $\begingroup$ @Eoin: Here's a question about the $99\%$ claim (and an answer that confirms it). $\endgroup$ – joriki Jul 29 '15 at 4:37