Understanding a group of order $2^{25}.97^2$ 
Let $G$ be a semidirect product of a Sylow 2-subgroup $P$ and a normal
  subgroup $Q$. 
$P$ is itself is semidirect product as defined below: $$P=(\langle u \rangle \times \langle v \rangle \times \langle w \rangle \times \langle h \rangle ) \rtimes (\langle a \rangle \times \langle g \rangle)$$ where $o(u)=32, o(v)=4, o(w)=8, o(h)=256,o(a)=64, o(g)=2$, with action defined as: 
  
  
*
  
*$u\in Z(P)$
  
*$v^a=u^{16}v, h^a=h^5, $ and $w^a=u^4w$
  
*$v^g=v^{-1}u^{-8}$, $[w,g]=[h,g]=1$
  

So $P$ is a group of order $2^{25}$. 
Now it can be verified that $Z(P)=\langle u,h^{64} \rangle$. Let $K=\langle u,v,g,h,a^8w \rangle$. We can check that $K\unlhd P$ and $P/K=\langle \bar{a} \rangle \cong C_{64}$ as $w^{-1}=a^8$ in $P/K$. 
Let $Q=\mathbb{F}_{97} \oplus \mathbb{F}_{97}$ as additive groups. The group $G=Q\rtimes P$ with $[Q,K]=1$ and $a$ acts faithfully on $Q$ by multiplication with the matrix $A=\begin{bmatrix}
0 & 1\\
\zeta & 0\\
\end{bmatrix}$, where $\zeta$ is (fixed) primitive $32nd$ root of unity of $\mathbb{F}_{97}$. One can verify that $G$ is metabelian with an abelian normal subgroup namely $\langle Q,v^2,u^4,h^4 \rangle$. Order of $G$ is $2^{25}.97^2$.
I am trying to understand this counterexample for Normalizer problem. It was published in Hertweck's paper but I am unable to deduce how he said $G$ is metabelian with $\langle Q,v^2,u^4,h^4 \rangle$ as normal abelian subgroup.
First claim he made was $Z(P)=\langle u,h^{64} \rangle$. Clearly $u$ is in centre because of first point. Also $h^{64}$ commutes with $u$ and $h$. But how is it concluded that $h^{64}$ commutes with $v$ and $w$. I cant see it from the second point which should be the reason somehow. 
Second how to prove that $K$ is normal and $P/K\cong <\bar{a}>$. And third is that why is it that $\langle Q,v^2,u^4,h^4 \rangle$ is abelian. 
Any help with this counterexample is much appreciated. I really have to understand it. Please suggest valuable ideas on this so I can make progress here. Thanks!
 A: 
First claim he made was $Z(P)=⟨u,h^{64}⟩$. Clearly $u$ is in centre because of first point. Also $h^{64}$ commutes with $u$ and $h$. But how is it concluded that $h^{64}$ commutes with $v$ and $w$. I cant see it from the second point which should be the reason somehow.

Note that $(h^{64})^a=h^{320}=h^{64}$. $h$ commutes with $v$ and $w$ by the definition of a direct product!

Second how to prove that $K$ is normal and $P/K\cong <\bar{a}>$. And third is that why is it that $⟨Q,v^2,u^4,h^4⟩$ is abelian. 

The second and third bullet points imply the subgroup $H=\langle u,v,h\rangle$ is normal. Note that in $P/H$, $a^8w$ and $g$ are central by the same bullet points (in fact, $P/H$ is abelian). So the preimage (which is $K$) is normal. The image is generated by $a$ because the image of $w$ in $P/K$ is $a^{-8}$, by the definition of $K$, and all other generators are already in $K$.
Finally, $⟨Q,v^2,u^4,h^4⟩$ is abelian because:


*

*$Q$ is

*$[Q,K]=1$ by assumption, and $(v^2,u^4,h^4)\subset K$

*$u$, $v$, and $h$ all commute

