I am interested in the following:


Let $G$ be a finite group of order $n$. Is there an explicit function $f$ such that 
$|s(G)| \leq f(n)$ for all $G$ and for all natural numbers $n$, where $s(G)$ denotes
the set of subgroups of $G$?



General bound for the number of subgroups of a finite group
A non-identity subgroup of a group of order $n$ can be generated by $\log\_{2}(n)$ or fewer elements. There is quite a lot of duplication, but if you count the number of subsets of $G$ of cardinality at most $\log\_{2}(n),$ you will have an upper bound (which could be made more precise with more care) and the case of elementary Abelian $2$-groups show that such a bound is of the right general shape if it is to cover all groups of all orders.


To be more precise, this gives an upper bound of approximately $\log\_{2}(n)n^{\log\_{2}(n)}$ for the total number of subgroups of a group of order $n,$ which is generically rather generous. On the other hand, the number of subgroups of an elementary Abelian group of order $n= 2^{r}$ is close to $\sum\_{i=0}^{r}2^{i(r-i)},$ so generally larger that $n^{log\_{2}(n)/4}.$


