Order of some subgroup of $\mathit{SL}_2(\mathbb Z/p^n\mathbb Z)$ I am trying to this problem from a past exam:

Let $p$ be an odd prime and $n$ a natural number.  Show that the group $G:= \left\{\begin{pmatrix}a& b\\0 & d\end{pmatrix}\right\} < \mathit{SL}_2(\mathbb Z/p^n\mathbb Z)$ has a unique subgroup of order $p^{2n-1}$.

I tried solving it in the following way:  To apply Sylow's theorem, I calculated the order of $G$. $\begin{pmatrix}a& b\\0 & d\end{pmatrix}$ is in $G$ iff $ad = 1$. Since $\mathrm{gcd}(a,p^n) = 1$ iff there exists a unique $d\in\mathbb Z/p^n\mathbb Z$ s.t. $ad=1$, $a$ can be an element of $\mathbb Z/p^n\mathbb Z$ that is coprime to $p^n$ and there exist a unique $d$ for each value of such $a$, if $\begin{pmatrix}a& b\\0 & d\end{pmatrix}$ is in $G$.  The number of such pairs $(a, d)$ is $p^n - p^{n-1}$.  And since $b$ can be chosen arbitrarily, $|G| = p^n(p^n - p^{n-1}) = p^{2n-1}(p-1)$. Then, since $p \not\mid p - 1$, one can apply Sylow's theorem to obtain the desired result.
My question is whether this is correct.  I doubt if it is, because I did not use the fact that $p$ is odd.

I believe the reason $p=2$ is omitted is because that case is trivial in comparison; $G$ would be a $2$-group so uniqueness is an immediate fact requiring no Sylow theory at all, only size checking.
Rather than say "since $p\nmid p-1$" I'd say "since $p-1$'s only divisor $\equiv1$ mod $p$ is $1$": other than this wording your argument appears to be fine to me.
