for $p>2$, If an element $x$ centrilize $\Omega(A)$ where $A\leq P$

Question

Let $P$ be a $p$ group for a prime $p>2$ and $A\leq P$ and $x\in C_P(\Omega_1(A))$.

Does it follows that $x\in N_G(A)$? If it not true, is it true when $A$ is abelian?

I was reading a proof of a theorem and I could not understand a part of it. I guessed that such a statement is true.

Note: $\Omega_1(A)=\langle \{x\in A \mid x^p=1 \}\rangle$

Answer 1

Let $p$ be a prime and $G$ the group defined by the presentation $$G=\langle x,y,z \mid x^{p^2}=y^p=z^p=1,xy=yx,z^{-1}xz=xy,zy=yz \rangle.$$ So $G$ has order $p^4$ and is a semidirect product of $\langle x,y \rangle = C_{p^2} \times C_p$ by $\langle z \rangle \cong C_p$.

Let $A=\langle x \rangle$. Then $z$ centralizes $\Omega_1(A) = \langle x^3 \rangle$, but does not normalize $A$.