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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.

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1 Answer 1

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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.

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