Let $G$ be a finite group. A chain of subgroups of $G$ of length $d$ is a sequence of subgroups of the form
$$ \{e\}=G\_0 \subsetneq G\_1 \subsetneq \ldots \subsetneq G\_{d-1} \subsetneq G\_d=G. $$
The length $l(G)$ of $G$ is, by definition, the length of the longest chain of subgroups of $G$.


Let $\Lambda(G)$ be the number of conjugacy classes of subgroups of $G$.


Obviously, one can bound $\Lambda(G)$ using the order $|G|$ of $G$. My question is whether one can bound $\Lambda(G)$ using $l(G)$. 


To be more precise: 


Fix an integer $n$. Does there exist, for any integer $m$, a finite group $G\_m$ such that $l(G\_m)\leq n$, but $\Lambda(G\_m)\geq m$?


Thanks.



length of a finite group versus number of conjugacy classes of subgroups
Let $G=(\mathbb{Z}/p\mathbb{Z})^n$, for $p$ a prime and $n>1$.


Then $l(G)=n$ but $\Lambda(G)$ can be made as large as you like by choosing $p$ large.


