Is there a formula for finding the number of groups of order n. For example, if a group $G$ has an order n, is there a formula in which someone can find the number of groups with that order. I suppose this would act similar to prime factorization since prime groups have only 1 group.
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1$\begingroup$ I believe that this is a very difficult problem in general. However, there is an exact answer for finite abelian groups. Let $\tau(n)$ represent the number of integer partitions of $n \in \mathbb{N}$. By the fundamental theorem of finitely generated abelian groups, the number of groups of order $p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\cdots p_{k}^{\alpha_{k}}$ is given by $\tau(\alpha_{1})\tau(\alpha_{2})\cdots \tau(\alpha_{k})$. $\endgroup$– Alex WertheimCommented Nov 21, 2014 at 20:53
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$\begingroup$ From there would using conjugacy classes be of any help? $\endgroup$– TeeJ LockwoodCommented Nov 21, 2014 at 20:56
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$\begingroup$ No such formula is known. There are some special cases where you have a formula (eg. for the number of groups of squarefree order). But currently there is no reason to expect that it will ever be easy to compute the number of groups of order $n$ (for arbitrary $n$). For example, nobody knows how many groups of order $2048$ exist. $\endgroup$– Mikko KorhonenCommented Nov 21, 2014 at 20:58
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4$\begingroup$ Congratulations! This is the first of all the sequences in the Online Encyclopedia of Integer Sequences! A000001 $\endgroup$– Hagen von EitzenCommented Nov 21, 2014 at 21:02
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$\begingroup$ Look at the tables that exist - the numbers of groups of prime power order grow very quickly (the prime $2$ is the best known, but not wholly known). The groups of prime power order have good properties (are nilpotent, for example), but the best results are for orders where the number of groups is small rather than where the number of groups is large. Solve this and you go down in mathematical history. $\endgroup$– Mark BennetCommented Nov 21, 2014 at 21:14
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2 Answers
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In general there is no formula $f(n)$ for the number of groups of order $n$ up to isomorphism. However, if $n$ is squarefree (no prime to the power $2$ divides $n$), then Otto Hölder in 1893 proved the following amazing formula. $$f(n)=\sum_{m \mid n}\prod_p \frac{p^{c(p)}-1}{p-1}$$ where $p$ is a prime divisor of $n/m$ and $c(p)$ is the number of prime divisors $q$ of $m$ that satisfy $q \equiv 1$ (mod $p$). From this formula it follows for example that $f(n)=1$ if and only if gcd$(n,\varphi(n))=1$, where $\varphi$ is the Euler totient function.
Recommended further reading is the paper of J.H. Conway, H. Dietrich and E.A. O'Brien - Counting groups:gnu's, moas and other exotica, (google this title and find the .pdf!), where you will encounter more formula's and also a table of the values of $f(n)$ up to $n=2048$. You will note that for $n$ is a prime power, $f(n)$ becomes very very large.
answered Nov 22, 2014 at 10:38
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I think number of groups of given order is infinite but number of groups of given order up to isomorphism is unknown.
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$\begingroup$ Clearly the OP meant up to IM. Otherwise trivially substituting one object for another changes the group. $\endgroup$ Commented Nov 21, 2014 at 21:03