About subgroups of the order of the normalizer of a p-subgroup

group-theorysylow-theory

I recently came across a doubt I've already partially exposed in this post without getting a solution. So I'd like to isolate my final concern from the textbook I mentioned there.
The real doubt is:

Is true that in a group $G$, if $P$ is a Sylow $p$-subgroup and $N$ is the normalizer of $P$, then the only subgroups of order $|N|$ are the normalizers of the Sylow p-subgroups?

I've checked within $A_5$ and that seems true in that case:

  1. there $6$ 5-sylow, with their related $6$ normalizers of order $10$, which are exhausting all subgroups of order $10$ in $A_5$
  2. there are $10$ 3-sylow, with their related $10$ normalizers of order 6, which are exhausting all subgroups of order $6$ in $A_5$
  3. there are $5$ 2-sylow of order $4$, which their related $5$ normalizers of order $12$, which are exhausting all subgroups of order $12$ in $A_5$

My doubt is this holds true for any generic group $G$ or just for $A_5$.

Thanks in advance.

Best Answer

No, a counterexample is $\mathtt{SmallGroup}(324,160)$, which has the structure $3^3\!:\!A_4$. It has a unique minimal normal subgroup $K$ of order $3^3$ and $G/K \cong A_4$.

The normalizer $N$ of a Sylow $2$-subgroup has order $12$ with $N \cong A_4$, but there is another conjugacy class of subgroups of order $12$, which do not have a normal Sylow $2$-subgroup.