Bound on the height of the subgroup lattice The height of the subgroup lattice is
$$h(G) = \max \Big\{N \in \Bbb N : \exists\ H_1, \ldots, H_N \leq G \text{ s.t. } \{e\} < H_1 < \ldots < H_N < G\Big\}$$
We can bound this by the order of $G$, $\mathcal{h}(G) < |G|$. 
Are there any better bounds available on the height of subgroup lattice, for arbitrary groups or particular families? 
 A: If you know the prime decomposition of the order of $G$, then you can get upperbounds specific to $G$.
Indeed if $|G| = p^{\alpha_1}_1 \cdots p^{\alpha_n}_n$ where $p_k$ are distinct primes and $\alpha_k \in \Bbb Z_+$, then the maximum possible length of a chain of subgroups takes a very simple form:
$$
h(G) \leq \boxed{\alpha_1 + \cdots + \alpha_n}
$$
The Reason: Note that the order of the subgroup $H_N$ must have the form $p^{\beta_1}_1 \cdots p^{\beta_n}_n$ (by Lagrange) where $\Bbb N \ni \beta_k \leq \alpha_k$ and at least one of the $\beta_k$ is strictly less than $\alpha_k$. The last condition is due to the fact that you want $H_N$ to be a strict subgroup of $G$.
An exactly analogous argument holds true with regards to the order of $H_i$ in comparison to the order of $H_{i+1}$ since each $H_i$ is also a strict subgroup of $H_{i+1}$.
Hence in the best case scenario a chain of subgroups $G > H_N > \cdots > H_1 > \{e\}$ will descend in order by decreasing one of the exponents $\alpha_k$ by just $1$ at each descent (if it decreased by more, you will reach $\{e\}$ faster and there will be fewer subgroups in the chain).
E.g. if $|G| = 7^3 \cdot 11^2$, then in one of the best case scenarios the orders in a chain of subgroups will descend like so:
$$
|G| = 7^3 \cdot 11^2 > 7^2 \cdot 11^2 > 7^1 \cdot 11^2 > 11^2 > 11^1 > 1 = |e|
$$
Of course there is a reason I said "one of the best case scenarios". Because there may be multiple ways to descend to $1$. Like in the example above, another possibility is
$$
|G| = 7^3 \cdot 11^2 > 7^3 \cdot 11^1 > 7^3 > 7^2 > 7^1 > 1 = |e|
$$
But the maximum possible length of any such descent remains the same.
Also I would like to point out that the maximum possible lengths are always achieved in cyclic groups because cyclic groups are guaranteed to have subgroups corresponding to every divisor of the order of the group.
So again using our example above: if $G$ were the cyclic group $\Bbb Z_{7^3 \cdot 11^2}$ with order $7^3 \cdot 11^2$, then here is a descending chain of subgroups that achieves the maximum length:
$$
G = \langle 1 \rangle > \langle 7^1 \rangle > \langle 7^2 \rangle > \langle 7^3 \rangle > \langle 7^3 \cdot 11^1 \rangle > \langle 0 \rangle = \{e\}
$$
