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1. Let $G$ be a finite group of order $n$. I need an estimate for the number $c$ of conjugacy classes of subgroups $D\subseteq G$.

Note that any subgroup of $G$ contains $1_G$, and so the set of all subgroups of $G$ is contained in the set of all subsets of $G$ containing $1_G$; this set of subsets has cardinality $2^{n-1}$. We obtain that $$ c\le 2^{n-1}.$$

Question 1. What are better estimates of $c$ in terms of $n$?

In Question 1, when counting the conjugacy classes of subgroups, I don't want to count the conjugacy class of a cyclic subgroup that is contained in a strictly larger cyclic subgroup. This is related to the Chebotarev density theorem; see Question 2.

2. In my application, $G={\rm Gal}(L/K)$, the Galois group of a finite Galois extension of global fields $L/K$. Then $G$ acts on $L$ and on the set of places of $L$. For any place $w$ of $L$, consider the decomposition group $D_w$, that is, the stabilizer of $w$ in $G$. We say that $D_w$ is a maximal decomposition group if it is not contained in a strictly larger decomposition group. Let $m$ denote the cardinality of the set of conjugacy classes of maximal decomposition groups. Clearly, we have $$m\le c\le 2^{n-1}.$$

Question 2. What are better estimates of $m$ in terms of $n$?

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    $\begingroup$ You probably can't do better than $n^{c\log{n}}$ for some constant $c$, as an elementary Abelian $2$-group of order $n = 2^{r}$ illustrates. I think I have answered questions like this before, but I could not find them. I think L. Pyber might have shown that this sort of bound is asymptotically optimal. By the way, the difference between "number of subgroups" and "number of conjugacy class of subgroups" is probably negligible, since they differ by a factor at most $n$. $\endgroup$
    – Geoff Robinson
    Commented Mar 18 at 14:18
  • $\begingroup$ Thank you for your prompt comment, @GeoffRobinson! A stupid question: what is the number of subgroups for the elementary Abelian $2$-group of order $n=2^r$ ? $\endgroup$
    – Mikhail Borovoi
    Commented Mar 18 at 14:34
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    $\begingroup$ The relevant question is mathoverflow.net/questions/132675 . The answers/comments there include references to Pyber's result. $\endgroup$
    – Geoff Robinson
    Commented Mar 18 at 15:09
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    $\begingroup$ The theorem mentioned in that answer is by Borovik, Pyber and Shalev. My own answer to that question is more informative than my comments above. $\endgroup$
    – Geoff Robinson
    Commented Mar 18 at 15:14
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    $\begingroup$ @GeoffRobinson: Many-many thanks! $\endgroup$
    – Mikhail Borovoi
    Commented Mar 18 at 15:20

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