Let $n$ be a prime, compute order of each element of each element of
$$G= \{ a,d : a^{2n}=e, d^2 =a^n,da =a^{-1}d $$
what is possible order of each element ?
number of elements of that order
the elements of that order
Progress?
$$G = \{ a^i d^j: 0 \leq i \leq 2n-1,0 \leq k \leq 1 \} $$
breaking it down as
$$\begin{matrix} a^0 d^0 & a^1 d^0 \dots a^{2n-1} \\ a^0 d^1 & a^1 d^1 \dots a^{2n-1} d \end{matrix} $$
Grabbing elment from the second row and raising to the second power
$$ (a^i d)(a^id)= a^i d a^i d= a^i a^{-i} dd= d^2$$
so raising it to the 4th power
$$(a^i d)^4= (a^i d)^2 (a^i d)^2=d^2 d^2 = a^n a^n=a^{2n}=e $$
grabbing elements of first row
$$ \begin{aligned} |a^1|=2p \\|a^2|=p \\ |a^3|=2p \\ |a^4|=p \\ |a^5|=2p \\ \vdots \end{aligned} $$
Looking at cases
$1$ no prime since unit consider $p=2$
$$ \begin{aligned} |a^1|=2p \\|a^2|=p \\ |a^3|=2p \end{aligned} $$
considre $p=3$
$$ \begin{aligned} |a^1|=2p \\|a^2|=p \\ |a^3|=2p \\ |a^4|=p \\ |a^5|=2p \end{aligned} $$
consider p=5
$$ \begin{aligned} |a^1|=2p \\|a^2|=p \\ |a^3|=2p \\ |a^4|=p \\ |a^5|=2p \\ |a^6|=p \\ |a^7|=2p \\ |a^8|=p \\ |a^9|=2p \end{aligned} $$
the pattern is that there are p of order $2p$ and $p-1$ of order $p$
But this is not technically a proof. Want to do something with induction but not sure how to properly word it.
Clearly, $|a| = 2n.$
Consider $|a^i|$.
We need values $k,l$ such that ${(a^i)}^k$ = $(a^{2n})^l$, i.e. $ik = 2nl$. This happens at $\text{LCM}(i,2n)$.
However $n$ is prime, so by dividing through by $i$ we get $k$ is the pattern you suggest.
I'll let you calculate the result for elements involving $d$ on your own.