Let $P$ be a finite nilpotent group.
We want to show that every proper subgroup of $P$ is a proper subgroup of its normalizer in $P$.
I was following the solution. But there are certain parts I didn't understand.
Suppose $H$ is a proper subgroup of $P$.
We know that it's always the case that if $H$ is a subgroup of $P$, then $H \le N_p(H)$.
We will show that if $H \ne P$, then $H \ne N_p(H)$.
Since $P$ is nilpotent, by definition using upper central series, $Z_i(P)=P$ for some $i \in \mathbb{Z}$.
Since $H$ is proper, $Z_{i-1}(P) \subseteq H \subsetneq Z_i(P)$.
Let $x\in Z_i(P) - H$. We will show that $x\in N_p(H)$.
Let $h \in H$. Then we have $[x,h^{-1}] = x^{-1}hxh^{-1} \in Z_{i-1}(P) \subseteq H$.
Thus, $x^{-1}hx \in H$.
Now here are my questions:
- Why it's that case that $Z_{i-1}(P) \subseteq H \subsetneq Z_i(P)$? I feel like that this has something to do with this theorem below.
But I still don't really see why $Z_{i-1} \subseteq H \subsetneq Z_i(P)$, where we can't have something like $Z_{i-2}(P) \subseteq H \subseteq Z_{i-1}(P) \subsetneq Z_i(P)$.
- Why do we have $x^{-1}hxh^{-1} \in Z_{i-1}(P)$?
Any help is appreciated!
By definition we have $1= Z_0(P)\le H$. Hence, there exists a largest $i\ge 1$ such that $Z_{i-1}(P)\le H$. Note that we do not necessarily have $Z_i(P)=P$ (I don't know where you got this from). Take $x\in Z_i(P)\setminus H$. By definition $x Z_{i-1}(P)\in Z(P/Z_{i-1}(P))$. It follows that $xHx^{-1}/Z_{i-1}(P)=H/Z_{i-1}(P)$ and $xHx^{-1}=H$. Therefore, $x\in N_P(H)$.