TITLE: General bound for the number of subgroups of a finite group
QUESTION [14 upvotes]: 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 $G$?

REPLY [3 votes]: In a similar vein to Geoff Robinson's answer, observe that any proper subgroup of a group of order $n$ can be generated by at most $\Omega(n) - 1$ elements, where $\Omega$ counts the number of prime factors (with multiplicity) of $n$. This is tight, with equality in the case of a product of prime cyclic groups.
This gives an upper bound of:
$$ \binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{\Omega(n) - 1} + 1$$
subgroups, where the final $+ 1$ includes $G$ itself. Now we can invoke Michael Lugo's strict upper bound from https://mathoverflow.net/a/17236/39521 to obtain a closed-form upper bound:
$$ |s(G)| \leq \left\lceil \binom{n}{\Omega(n) - 1} \dfrac{n - \Omega(n) + 2}{n - 2 \Omega(n) + 2} \right\rceil \leq \dfrac{2 \times n^{\Omega(n) - 1}}{(\Omega(n) - 1)!}$$
Note that this is tight infinitely often: in the case where $n = p$ is prime (so $G = C_p$), we get:
$$ \dfrac{2 \times p^0}{0!} = 2 $$
where the two subgroups are the trivial group and $G$ itself.