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For a my class in Abstract Algebra, our teacher asked us to answer the question above, noting it's a surprisingly large number.

I have found this question More than 99% of groups of order less than 2000 are of order 1024?. It seems like the answer should be even greater when looking at groups of order less than 10,000 but I would not know how to prove it or state it.

Now, I think the number 10,000 was arbitrary, used as an example, what my teacher is actually looking for is an answer like "99%" or "almost all". Still, my problem persists - I have no idea how to prove this. A hunch tells me that it should boil down to the fact that 2 is a prime factor of so many numbers (all the evens).

Any help is appreciated.

D. Brito
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  • Chickenmancer's answer suggests that the number (10,000) is very important. For the analysis of groups of order at most 100, the 51 groups of order 32 give rise to at least 51 groups of order 323 by taking a direct product with $C_3$. However, for groups of order at most 10,000, you can only use this trick to say that there are at least as many groups of order $2^{11}3$ as there are of order $2^{11}$. Assuming there are many more groups of order $2^{12}$ and $2^{13}$, this suggests that much more than one third are 2-groups. – Robert Bell May 12 '18 at 02:20
  • @RobertBell: Interesting observation. We could generalize this, motivated by the following example: $2^{10}3<2^{10}5<2^{13},$ then we end up counting the number of groups of order $2^{10}$ for each of these primes, and then again for products, (assuming that the product doesn't exceed the given restriction on group order). – Chickenmancer May 12 '18 at 03:40

1 Answers1

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The maximal $2-$group in this situation is of order $2^{13}.$ Now, the question comes down to deciding how many non-isomorphic $2-$groups exist with order less than or equal to $2^{13},$ and comparing that to the totality of groups of order less than $10,000.$ I can't speak to what these numbers should be . . . more funding is needed.

Just looking at the non-isomorphic groups of order less than or equal to $100$ is rather taxing. There are 1048, see:

https://www.mimuw.edu.pl/~zbimar/small_groups.pdf

Now, in terms of $2-$groups, there are relatively few. The total number of non-isomorphic $2-$groups of order $64$ is $267$, of order $32$ is $51$, of order $16$ is 14, or order $8$ is $5$, of order $4$ is $2,$ and of order $2$ is $1.$ In total

$$\text{total number of $2-$groups}= 267 + 51 +16 +5+2+1=342.$$

(http://oeis.org/A000679)

So we get $342/1048\approx 32\%.$ Rober Bell's comment, gives an idea of why the number of 2-groups of a given order is essential to the number of groups less or equal to a given order.

This can be quite difficult to compute, as the number of non-isomorphic groups with order $1024$ is $49,487,365,422$ . . .

(https://mathoverflow.net/questions/46855/how-to-generate-all-finite-groups-of-order-n)

Suppose we know the number of $2-$groups less than a given order $N.$ And let $n$ be the maximal value so that $2^n\leq N.$ We can give $N$ a prime factorization $$N=2^n(p_1^{e_1}\cdots p_m^{e_m}).$$ Suppose we know the total number of $2-$groups of orders $2,2^2,\ldots,2^n.$ Let us denote these as $m_1,\ldots,m_n$ respectively. Then for each $p_i$ we have at least

$m_kp_i^\ell$ groups of order $2^kp_i^\ell,$ for $1\leq k\leq n,$ $1\leq \ell\leq e_i.$ Hopefully this gives the idea of how one might compute the number of groups, up to isomorphism, less than a given order.

Chickenmancer
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  • What do you mean by over counting? – Tobias Kildetoft May 12 '18 at 05:54
  • I suppose over counting isn't quite correct. The idea is that if you have a $2-$group, then taking the direct (semi direct) product of the $2-$group and a group of relatively prime order will produce a non-isomorphic group, for each $2-$group. – Chickenmancer May 12 '18 at 06:00