I am trying to learn the proof techniques of group theory, and I tried to prove the following theorem (with some help from the book). Could you please tell me if my proof is correct, and give any suggestions or general comments? Thank you very much.
I have colorized in red the parts where I am slightly shaky. Thanks
Let $P$ be a $p$- group (so $|P| = p^a$ for some $a > 1$) and suppose $H < P$. Then $H<N_P(H)$.
Proof: We use induction on the order of $P$. The theorem is clearly true for $|P| = p^1$. Suppose the theorem is true for $|P| = p^1, p^2, ..., p^{a-1}$. Take a group $P$ of order $p^a$, and take $H < P$. If $Z(P) \not \subseteq H$, then we are done, because $Z(P) \le N_P(H)$. Suppose $Z(P) \subseteq H$, and consider the quotient group $\overline{P} = P / Z(P)$. Since (as I have proved before) the center of $P$ is not trivial, this quotient group is a $p$-group of order strictly less than $p^a$, and therefore it satisfies the induction hypothesis. Thus there exists an $\overline{x} \in \overline {P}$ with $\overline{x} \not \in \overline H$ such that $$\overline{x} \overline {H} \overline {x^{-1}} = \overline{H}.$$ $\color{red}{\text{The condition } \overline {x} \not \in \overline{H} \text{implies that } x \not \in H}$.
Take any $h \in H$. We have $\overline {x} \overline {h} \overline{x^{-1}} = \overline{h'}$ for some $h' \in H$, $\color{red}{\text{which implies that } xhx^{-1} \in H}$.
Since $h$ was arbitrary above, we have that $xhx^{-1} \in H$ for all $h \in H$, and therefore $x \in N_P(H)$ but $x \not \in H$, as desired.