Normalizer of dihedral group in projective linear group I wish to prove that $N_{PGL(2,p^2)}(D_{2p})\cong Z_p \rtimes Z_{p-1}$given $p$ an odd prime, ($|D_{2p}|=2p$ here) . ALso, instead of 2, one can generalize the power to be any $k$ where $p^k -1 $ divisible by 4. 
Any guidance or sketch of proof is tremendously valued.  
 A: As I said in my comment (which I have deleted because it had a mistake), the normalizer of  $D_{2p}$ in ${\rm PGL}(2,p^2)$ is contained in its normalizer in the affine group $A \rtimes_{\phi} B$, where $A$ is the additive group $({\mathbb F}_{p^2},+)$, $B$ is the multiplicative group $({\mathbb F}_{p^2}^*,\times)$,and the action $\phi$ is given by field multiplication. In this notation,the $D_{2p}$ subgroup can be taken to be $\{ (x,y) : x \in {\mathbb F}_p, y = \pm 1 \}$.
Suppose that $(a,b)$, with $a \in A$ and $b \in B$ is in the normalizer of this subgroup. Then $(a,b)(0,-1) = (x,y)(a,b)$ for some $(x,y) \in D_{2p}$, giving $(a,-b) = (x+ya,by)$, so $y= -1$, and hence $a=x-a$, so $a = x/2 \in {\mathbb F}_p$.
Also $(0,-1)(a,b) = (a,b)(x,y)$ for some $(x,y) \in D_{2p}$, giving $(-a,-b)= (a+bx,by)$, so $2a=bx$ and either $x=a=0$ or $b = 2a/x \in \mathbb{F}_p^*$.
If $a=0$, then $(0,b)(1,1) = (x,y)(0,b)$ for some $(x,y) \in D_{2p}$, giving $(b,b) = (x,yb)$ and again $b = x \in \mathbb{F}_p^*$.
