# Bounds for conjugacy classes of subgroups

My question is rather general and the reason for this is that I am primarily interested in finding sources where I can read more about this type of problems. So here goes:

To begin with, I want to find out more about methods to determine the number of conjugacy classes of subgroups of a given order in an ambient group $G$. A specific question I have in this vein is this:

Suppose $G$ is a finite 2-group and $N \triangleleft G$ is a normal subgroup of $G$ such that $G/N \cong N$. Now assume that $N$ has $f(i)$ conjugacy classes of subgroups of order $2^i$, $i \in \{0,1, \dots , \log_2|N|\}$. Is it true that $G$ has (at most) $f(i)f(j)$ conjugacy classes of subgroups of order $2^{i+j}$?

If the answer is "no" in general, is it possible to make further assumptions on either $G$ or $N$ in order to obtain an affirmative answer?

P.S. You may assume that $N=Z(G)$ if that will make the problem more interesting.

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## 1 Answer

If you take the Klein group $G$, of order $4$, and $N$ a subgroup of order $2$, then $f(0) = 1$, $f(1) = 1$. Take $i = 0, j = 1$, so that $f(0)f(1) = 1$. But in $G$ you have $3$ (normal) subgroups of order $2 = s^{i+j}$.

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Ok, thanks. I'll make it a little more interesting then. – user13040 Feb 7 '13 at 19:59
Oh, ok @the_fox, I'll try to run GAP to get examples. – Andreas Caranti Feb 8 '13 at 17:15
@the_fox, it is trivial to find counterexamples when $i=0$ and $j=\log_2|N|$, but this extreme case can be excluded. I am looking at more interesting examples. – Andreas Caranti Feb 8 '13 at 18:28
Do you have a concrete counterexample when $i=0$ and $j=\log_2|N|$? – user13040 Feb 8 '13 at 19:22