# On the average number of subgroups per conjugacy class

At some point in the future, I hope to do some work on estimates for the number of conjugacy classes of subgroups of a finite group (by an estimate here, I mean an upper bound). Assuming, for the moment, that such an estimate has been proven, a natural next question is to ask for a bound for the total number of subgroups of a group.

So, if $G$ is a finite group, let $s(G)$ denote the number of subgroups of $G$, and let $c(G)$ denote the number of conjugacy classes of subgroups of $G$. Given $G$, define $r(G) := s(G)/c(G)$. Obviously, $r(G) \geq 1$ for each finite group $G$, and equality occurs if and only if $G$ is Dedekind, i.e., every subgroup of $G$ is normal in $G$.

Also, $r(G)$ is unbounded. For example, if $p$ is an odd prime, then the dihedral group $D_{2p}$ has exactly 4 conjugacy classes of subgroups, but $p+3$ subgroups overall. In fact, some initial experiments with GAP suggest that $D_{2p}$ is the "worst case", by which I mean that

$$r(G) \leq \frac{\frac{|G|}{2}+3}{4} = \frac{|G|+6}{8} \tag{*}\label{*}$$

holds for every finite group $G$ (with equality if and only if $G \cong D_{2p}$).

It is easy to deduce an upper bound of the form $r(G) < |G|/q$, where $q$ is the smallest prime divisor of $|G|$, but actually proving \eqref{*} seems to require some effort.

1. My first question is whether there is an argument to prove \eqref{*}.
2. My second question is whether it is possible to do substantially better than "linear in terms of the order of the group" for the upper bound. Most groups (of small order) that I have checked have very low $r$'s; usually below 3. Which parameters of $G$ control the behaviour of $r(G)$?

There are quite a few papers by Brandl, Fernández-Alcober, La Haye, Legarreta, Poland, Rhemtulla, where a related quantity is studied. For $G$ a finite group, write $\nu(G)$ for the number of conjugacy classes of non-normal subgroups of $G$. Subsets of these authors study $\nu(G)$, and usually focus on the case where $G$ is a $p$-group.

Of the more interesting results that have been obtained in this line is that if $G$ is a $p$-group and $\nu(G)>0$, then $\nu(G)=1$ or $\nu(G) \geq p$. Brandl has described explicitly the $p$-groups which realise $\nu(G) \leq p+1$. Some relevant literature follows.

• R. Brandl, Groups with few non-normal subgroups, Comm. Algebra 23 (6) (1995) 2091–2098.

• J. Poland, A. Rhemtulla, The number of conjugacy classes of non-normal subgroups in nilpotent  groups, Comm. Algebra 24 (10) (1996) 3237–3245.

• R. La Haye, A. Rhemtulla, Groups with a bounded number of conjugacy classes of non-normal subgroups, J. Algebra 214 (1999) 41–63.

• G.A. Fernández-Alcober, L. Legarreta, Conjugacy classes of non-normal subgroups in finite nilpotent groups, J. Group Theory 11 (2008) 381–397.

• G.A. Fernández-Alcober, L. Legarreta, Bounds for the number of conjugacy classes of non-normal subgroups in a finite p-group, Comm. Algebra 37 (11) (2009) 3928-3942.

• G.A. Fernández-Alcober, L. Legarreta, The finite p-groups with p conjugacy classes of non-normal subgroups, Israel J. Math. 180 (2010) 189–192.

• R. Brandl, Conjugacy classes of non-normal subgroups of finite p-groups, Israel J. Math. 195 (2013) 473–479.

• mathoverflow.net/questions/132675/… might be relevant. – Benjamin Steinberg Apr 25 '17 at 11:50
• I am not aware of any known results in this area, but my guess is that it would be possible to prove something similar to $(*)$ with some work. For $(*)$ to fail there would have to be subgroups with normalizers of size smaller than $8$, which is possible, but has consequences on the structure of $G$. For example, if there is a self-normalizing subgroup of order $2$ or $3$, then $G$ has a normal $2$- or $3$-complement by Burnside's Transfer Theorem. – Derek Holt Apr 25 '17 at 12:09
• On the other hand, it is possible that this problem has been studied already, so you should try and ascertain that before working too hard on it! – Derek Holt Apr 25 '17 at 12:11
• In the past, I had searched for results related to $s(G)$ or $c(G)$, but I did not find much of interest there... – the_fox Apr 25 '17 at 12:20
• Your suggestion to use Burnside is good; I hadn't thought of that. I think that it might be useful to try to answer the following: if $G=[N]H$ is a semidirect product of the normal $N$ with $H$, where $\gcd(|N|,|H|)=1$, is $r(G)$ related to $r(G/N)$, $r(H)$ in a "nice way"? – the_fox Apr 25 '17 at 12:38