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