# How many groups of order $2500$ are there?

I aborted the GAP-calculation of $Size(ConstructAllGroups(2500))$ after about $3$ hours. $gnu(2500)$ seems to be a very hard case.

Does anyone know $gnu(2500)$ (The number of groups of order $2500$), or at least whether it is smaller than , larger than or equal to $2500$ ?

The number of groups of order $d$ is smaller than $2500$ for every proper divisor $d|2500$, so $gnu(2500)$ could be smaller then $2500$.

• At least, there are only very few primitive permutation groups of order $2500$, see here, but of course, that does not really help. But the paper is interesting. Jan 8 '16 at 11:07

There are 227 groups (from a 5 minute calculation in GAP with minor improvements as outlined in previous answers).

• Is the improved version already available ? Jan 8 '16 at 20:16
• @Peter If you are willing to work with my version of the code (i.e. you need to pull the source and compile on your own) it is in github under github.com/hulpke/gap/tree/work . It should migrate over the next weeks into the main GAP github repository. If you are willing to edit your GAP installation I can also send you a patch. Jan 8 '16 at 21:12
• Well, the improved version is magnitudes faster than the one I have. Of course, I would like to work with it. What do I have to do exactly ? Jan 8 '16 at 21:14
• @Peter Willing to work'' means also to work with installing code off a repository. Jan 8 '16 at 21:15
• Then, I do not think I manage it :( I hoped that it would be sufficient to download another group-construction-directory. Jan 8 '16 at 21:17

Applying Sylow's theorem, there is a unique $5$-Sylow which is then normal in $G$ if $G$ is a group of order $2500$. We denote it $N$.

On the other hand, take $H$ to be one $2$-Sylow of $G$ (it is of order $4$).

It is clear that :

$$G=N\rtimes H$$

Furthermore (again by Sylow's theory) the isomorphism class of $N$ and $H$ are uniquely defined by $G$, in other words if $N'$ and $H'$ are groups of order $5^4$ and $4$ respectively then :

$$N\rtimes H\text{ is isomorphic to } N'\rtimes H'\implies N\text{ is isomorphic to } N'\text{ and }H\text{ is isomorphic to } H'$$

You have fifteen groups of order $p^4$ and all of them can be written as semi-direct product of cyclic groups (see theorem 1.6.1 and theorem 1.6.14 in https://people.kth.se/~boij/kandexjobbVT11/Material/pgroups.pdf) and two groups of order $4$.

If you fix $N$ and $H$ then to determine the number of $G$ which are isomorphic to some semi-direct product $N\rtimes H$, it suffices to compute $Hom(H,Aut(N))$. Of course different elements in $Hom(H,Aut(N))$ could lead to isomorphic group. After all, $Aut(N)$ is computable if you use GAP for each group $N$ of cardinal $5^4$. Then computing the cardinality of $Hom(H,Aut(N))$ may not be very difficult... I think you can work this out...