An upper bound for the number of semidirect products of a finite group up to isomorphism Given a group $G$ with identity element 1, a subgroup $H$, and a normal subgroup $N$ of $G$; $G$ is called the semidirect product of $N$ and $H$, written $G = N\rtimes H$ , if  $G = NH$ and $H\cap N=1$.  
My question is:  
How many semidirect products are there for a finite group $G$ up to ismomorphism? Is there any upper bound for it?
 A: There clearly is an upper bound. A particularly bad one would be the number of subgroups squared to estimate pairs $(N,H)$ of subgroups of $G$. Since subgroups are subsets you get at most
$$
2^{2\cdot|G|}
$$
subdirect products.
As for better counts there is ambiguity concerning what you mean by "isomorphism". Do you want to classify up to: (these are in general different equivalences, in the order given they become more encompassing):


*

*pairs $(N,H)$ of subgroups of $G$ up to Automorphisms of $G$? (In this case the action is automatically defined by the subgroups.)

*Equivalences of extensions? (I.e. you construct external semi direct products and need to consider all possible actions and check which ones result in groups isomorphic to $G$)

*Isomorphism types of the group $G$? (Well, that would be just one class)
There are further gradations in between, which in certain circumstances might be considered as appropriate equivalences.
You might want to investigate the group $C_2\times C_2=V_4$ (the non-cyclic group of order 4) to get an idea what these different classes of equivalences are.
