This post is a *relative version* of [General bound for the number of subgroups of a finite group](https://mathoverflow.net/q/132675/34538)


Let $[H,G]$ be a interval of finite groups with $|G:H| = n$.


*Question:* What is a good upper-bound of $|[H,G]|$, as a function of $n$? 


If $H=\{ e\}$, then the best possible upper-bound is essentially $n^{(\frac{1}{4}+o(1)) \log\_2(n)}$ ([here](http://www.ams.org/tran/1996-348-09/S0002-9947-96-01665-0/S0002-9947-96-01665-0.pdf), Corollary 1.6).  

Should we expect the same in general?


There is an OEIS page for the maximal cardinal of a subgroup lattice for a group of order $n$: [A018216](http://oeis.org/A018216)  

1, 2, 2, 5, 2, 6, 2, 16, 6, 8, 2, 16, 2, 10, 4, 67, 2, 28, 2, 22, 10, 14, 2, 54, 8, 16, 28, 28, 2, 28, 2, 374, 4, 20, 4, 78, 2, 22, 16, 76, 2, 36, 2, 40, 12, 26, 2, 236, 10, 64, 4, 46, 2, 212, 14, 98, 22, 32, 2, 80, 2, 34, 36, 2825, 4, 52, 2, 58, 4, 52, 2, 272 


I didn't find an OEIS page for the maximal cardinal of an interval of finite groups, at index $n$.  

This should be computable for the indices $<32$, using GAP or MAGMA. 


*Remark*: Because an element $K \in [H,G]$ admits a unique partition by cosets $Hg$, we have:  

$$|[H,G]| < \sum\_{k \mid n} {n \choose k}$$
See [this post](https://math.stackexchange.com/q/1808549/84284) for a discussion on the estimate of this sum.  

We have slightly better with $\sum\_{k>1, k|n}^n {n-1\choose k-1}$.



A good upper-bound for the cardinal of an interval of finite groups
A chain in $[H,G]$ has length $\leq\Omega(|G:H|)$, where $\Omega$ denotes the number of prime factors counted with multiplicity. If $H=H\_0<H\_1<\dots<H\_k=G$ is a maximal chain, then there are elements $g\_1, \ldots, g\_k$, such that $H\_i=\langle H\_{i-1}, g\_i\rangle$. If $g\_iH=g\_i'H$, then $\langle g\_i H\_{i-1}\rangle = \langle g'\_i H\_{i-1}\rangle$, hence, the number of subgroup chains is bounded above by the number of sequences of cosets of $H$ in $G$, of length $\leq\Omega(|G:H|)$. Hence for $n=|G:H|$ the number of subgroups is $\leq \Omega(n) \frac{n!}{(n-\Omega(n))!} $ $ \le\Omega(n) n^{\Omega(n)}$ $\leq \log\_2(n) n^{\log\_2(n)} $ $ \leq n^{1+\log\_2(n)}$.


To gain the missing factor 4 in the exponent you have to use the fact that there are different choices for the $g\_i$, and that a group $K$ can occur in many different chains. However, this will be quite difficult. If you look at $C\_p^n$, you see that most subgroups have index $\approx p^{n/2}$, thus a local argument will not work.


