Now, it is not quite true that $P_n$ has more divisors than any $N < P_n$. It is a good strategy for making a number with many divisors, but soon enough it is better to add more factors of $p_1$, then eventually more factors of $p_2$, etc., than to keep adding new prime factors. At To understand this situation better, we can make many numbers (but not all numbers) that have more divisors than their predecessors with the "threshold method". The idea is to optimize the ratio $\log(d(N))/\log(N)$ globally by optimizing it locally (with respect to prime factorization). Pick a constant $t > 0$, the threshold, and say that $N$ should have at least qualitatively$k > 0$ factors of a prime $p$ if and only if$$\frac{\log(k+1) - \log(k)}{\log(p)} \ge t.$$Then I think that $d(M) < d(N)$ when $M < N$.
In fact, finding large values of $cd(G)$ (which I will use to mean the same thing eventually happens for cardinality of the group character degrees rather than the set) is a very similar problem when $H_n$. G$is nilpotent. A finite group is nilpotent if and only if it is a the product of$p$-groups. You can make a nilpotent group that looks similar to a number with a record number its Sylow subgroups. The main idea of divisors, by slowly increasing the$p$-group factor for small values of construction above is that in this case$p$. In factcd(G)$ is multiplicative, a particularly good choice i.e., the product of its values for $p$-groups. Following the comment by Frieder Ladisch, $p$ cd(G)$is the maximized for$k p$-groups by$C_{p^m} \times k$upper-triangular group over ltimes C_{p^{m+1}}$. (In the first version of the answer I used other $\mathbb{Z}/p$. This p$-groups that aren't as good.) I.e., this group has dimension character degrees$p^{O(k^2)}$, 1,p,\ldots,p^m$, and it was shown by Isaacs that it has no $O(k^2)$ distinct dimensions of irreducible representations (p$group with a different constant factor)$p^{2m}$or fewer elements can have an irrep with$p^m$elements. So , if you regularize the divisor function can find many record values of$d(n)$by taking cd(G)$ for nilpotent groups using instead the smallest monotonic function threshold formula$d'(n)$\frac{\log(k+1) - \log(k)}{\min(4-k,2)\log(p)} \ge d(n)$, then it is conceptually similar to t.$$Let's incorporate the same regularization concept of a "record value" by defining d'(N) to be the lower boundmaximum of d(M) with M \le N. I'd have Likewise define cd'(N) to be the maximum of cd(G) with |G| \le N. Then I think about how far apart that the bounds above constructions show that d'(N) and cd'(N) are at least similar functions, thoughand that$$d'(N) > cd'(N) > \sqrt[3]{d'(N)}$$when N is large enough. In fact I might add think that the exponent of the second inequality climbs from 1/3 to some higher value, although for nilpotent groups one also has$$\sqrt{d'(N)} > cd'_{\text{nil}}(N).$$Let me also mention that the bound O(\sqrt[3]{|G|}) follows immediately from the fact that |G| is the sum of the squares of the dimensions of the irreducible representations --- maybe that's what you have in mind with your bound. 1 The number of divisors at least roughly resembles the best achievable lower bound. For each prime p, there is a group G_p with p^3 elements which has an irreducible representation of dimension 1 (the trivial representation) and an irreducible representation of dimension p (because it's non-abelian). Now let p_n be the nth prime. The number$$P_n = p_1p_2\cdots p_n$$is a type of number with a lot of divisors. If you likewise let$$H_n = G_{p_1} \times G_{p_2} \times \cdots \times G_{p_n},$$then$H_n$has an irreducible representation for every$d$that divides$P_n$, and it has$P^3_n$elements. Now, it is not quite true that$P_n$has more divisors than any$N < P_n$. It is a good strategy for making a number with many divisors, but soon enough it is better to add more factors of$p_1$, then eventually more factors of$p_2$, etc., than to keep adding new prime factors. At least qualitatively, the same thing eventually happens for the group$H_n$. A finite group is nilpotent if and only if it is a product of$p$-groups. You can make a nilpotent group that looks similar to a number with a record number of divisors, by slowly increasing the$p$-group factor for small values of$p$. In fact, a particularly good choice for the$p$is the$k \times k$upper-triangular group over$\mathbb{Z}/p$. This group has dimension$p^{O(k^2)}$, and it was shown by Isaacs that it has$O(k^2)$distinct dimensions of irreducible representations (with a different constant factor). So, if you regularize the divisor function$d(n)$by taking the smallest monotonic function$d'(n) \ge d(n)$, then it is conceptually similar to the same regularization of the lower bound. I'd have to think about how far apart the bounds are, though. I might add that the bound$O(\sqrt[3]{|G|})$follows immediately from the fact that$|G|\$ is the sum of the squares of the dimensions of the irreducible representations --- maybe that's what you have in mind with your bound.