1

I know that there exists a unique group of order $n$ if and only if $(n, \phi(n))=1$. I also know how to obtain the number of abelian groups of a certain order

There are exactly 5 groups of order 8, two of them are non-abelian and there are exactly 5 groups of order 12, three of them are non-abelian.

I noticed that $\phi(8)=4=(8,4)$, $\phi(12)=4=(12,4)$ and that $2=\frac{8}{4}$,$3= \frac{12}{4}$.

So, my question is are there exactly $\frac{n}{\phi(n)}$ or $\frac{n}{(\phi(n),n)}$ non abelian groups of order $n$? Or is there any way to know how many non-abelian groups of a certian order there are?

allizdog
  • 927
  • Your guess is already wrong for $n=6$. There are not $6/\phi(6)=3$ non-abelian groups of order $6$, but just one, namely $S_3$. The question has been asked here, too. – Dietrich Burde Dec 20 '16 at 22:00
  • Okay so my guess is wrong. In the question you refer me to, I can only find a table of how many non-abelian groups there exist, but they don't show how to obtain those numbers. – allizdog Dec 20 '16 at 22:19
  • 1
    @allizdog That's because those numbers were all obtained through some kind of brute force and exceedingly technical reasoning; counting isomorphism classes of groups is extremely hard and, nowadays, computer-assisted in a big way (see here for instance). – pjs36 Dec 20 '16 at 22:24
  • Okay, so brute force is the only way to proceed, thank you! – allizdog Dec 20 '16 at 22:26
  • See also the classification of finite simple groups: https://en.wikipedia.org/wiki/Classification_of_finite_simple_groups – ChocolateAndCheese Dec 21 '16 at 01:36

0 Answers0