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

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 $$A_5$$.

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.
• I was inspecting some candidates for the other subgroups of order 12. The dihedral D12 would have the same Sylow 2-subgroups as $A_4$, i.e. Klein four groups, without having them normal in D12. I will double check my Guess with SAGE. Aug 16 at 13:10
• The other subgroup is isomorphic to $D_{12} \cong C_2 \times D_{6}$. It has a normal subgroup of order $3$ that is contained in the minimal normal subgroup of $G$. Aug 16 at 17:15