What are the normal subgroups of $PSL\_2(\mathbb{Z}/p^n \mathbb{Z})$?



Normal subgroups of projective special linear group over a ring
Let's suppose that $p >3$ (otherwise, the groups is solvable in any case). I also work with
$G = {\rm SL}(2,\mathbb{Z}/p^n \mathbb{Z})$, but there is an obvious correspondence between
what happens for ${\rm PSL}$ and what happens for ${\rm SL}$. The group $G$ has a largest
normal $p$-subgroup (denoted, as is customary for group theorists, by $O\_p(G)$), and
a central subgroup $Z$ of order $2$. $G/Z \times O\_{p}(G)$ is isomorphic to the simple group ${\rm PSL}(2,\mathbb{Z}/p\mathbb{Z})$. Furthermore, $H = O\_{p}(G)$ is the set 
of elements of $G$ which are congruent (entrywise) to the identity $(mod p)$. More generally, we have a series of obvious normal subgroups 
$$ 1 = H\_n < H\_{n-1} < \ldots < H\_{1} = H < G,$$ where $H\_{i}$ is the set of elements of $G$ congruent (entrywise) to the identity $(mod p^i)$. For each $j$, $H\_{j}/H\_{j+1}$ is an elementary Abelian $p$-group (that is, an Abelian $p$-group of exponent $p$).
To see this, note that for $a,b \in H\_j$, we have $ab -ba = (a-I)(b-I) - (b-I)(a-I) \equiv 0$ (mod $p^{2j})$, so that $I - a^{-1}b^{-1}ab \equiv 0$ (mod $p^{2j}$). Hence 
$[H\_{j},H\_{j}] \subseteq H\_{2j} \subseteq H\_{j+1}$ and $H\_{J}/H\_{j+1}$ is Abelian
(in fact $H\_{j}/H\_{2j}$ is Abelian). Also , for $a \in H\_j$, we have 
$a^{p} - I = (a-I)(a^{p-1}+ \ldots a^{p-2} + \ldots + I) \equiv 0$ (mod $p^{j+1})$,
so that $a^p \in H\_{j+1}$. Each of the factor groups $H\_{j}/H\_{j+1}$ has the structure
of a module for the group ${\rm SL}(2,\mathbb{Z}/p\mathbb{Z})$ over the field of $p$-elements. 


