Number of all possible groups of given order [duplicate]

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 $n$?

marked as duplicate by Zev Chonoles, anon, joriki, Davide Giraudo, Derek Holt group-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 29 '15 at 9:17

• 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. – the_fox Jul 29 '15 at 4:06
• 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. – pjs36 Jul 29 '15 at 4:06
• @pjs36 I remember reading somewhere more than 99% groups of order $<2000$ are of order $1024$. – Eoin Jul 29 '15 at 4:19
• @Eoin: Here's a question about the $99\%$ claim (and an answer that confirms it). – joriki Jul 29 '15 at 4:37