Normal subgroup of classical groups For $r\leq n$, consider the following reduction homomorphism 
$$
\pi_{n,r}: {\rm SL}_2(\mathbb{Z}/(p^n\mathbb{Z}))\to {\rm SL}_2(\mathbb{Z}/(p^r\mathbb{Z})).
$$
Bourgain and Gamburd in their paper "Expansion and random walks in ${\rm SL}_d(\Bbb Z/(p^n
\Bbb Z))$" mentioned that the set 
$$\{\ker\pi_{n,r}\},$$
gives all normal subgroups of ${\rm SL}_2(\Bbb Z/(p^n\Bbb Z))$. Probably this is standard, but I am not able to prove it.
Can any one please give me a hint or a reference for this fact? 
I remember that I read somewhere that a modification of  this fact also valid for 
$${\rm SL}_k(\mathbb{Z}/(p^n\mathbb{Z})),$$ 
and the symplectic group 
$${\rm Sp}_{2k}(\mathbb{Z}/(p^n\mathbb{Z}).$$
But I can not find where I saw these. I would be most thankful if anyone can help me with these. 
 A: Let me answer the question for $p>3$ and $SL_2$. I imagine that a similar method will work for the other cases but I haven't checked.
Some notation: $G=SL_2(\mathbb{Z}/p^n\mathbb{Z})$, $Z=\{I, -I\}$ and, for $i=1,\dots, n$,
$$K_i := \ker \pi_{n,i}.$$
Proposition: The proper normal subgroups of $G$ are $K_i$ and $K_i\times Z$ for $i=1,\dots, n$.
Sketch of basic steps of proof:


*

*If $n=1$, $G$ is quasisimple and the result is immediate. Assume from here on that $n>1$.

*Observe that $K_1$ is a normal
$p$-subgroup of $G$, that
$|K_1|=p^{3(n-1)}$ and $G/K_1\cong
   SL_2(\mathbb{Z}/p\mathbb{Z})$, a
quasisimple group since $p>3$.

*Observe, next that, for $i=1,\dots,
   n$, |$K_i| = p^{3(n-i)}$ and,
moreover, $$\{1\}=K_n \lhd K_{n-1}
   \lhd \cdots \lhd K_1$$ is a chain of
normal subgroups. In fact this is an
upper central series for the group
$K_1$, i.e $K_{i-1}/K_i = Z(K_1/K_i)$
and $K_{i-1}/K_i$ is elementary
abelian of order $p^3$.

*Now $G$ acts naturally on the group
$K_1$ by conjugation. The upper
central series structure just
described implies that this action
induces an action of $G/K_1 =
   SL_2(\mathbb{Z}/p\mathbb{Z})$ on the
groups $K_{i-1}/K_i$. Thus the group
$K_{i-1}/K_i$ becomes a 3-dimensional
module for the group
$SL_2(\mathbb{Z}/p\mathbb{Z})$. It is
easy to see that for $i=2,\dots, n$,
these modules are isomorphic. Fact to
check: This module is irreducible.

*Now let $N$ be a normal subgroup of
$G$. Suppose first that $N\cap K_1$
is trivial. Then $N$ is isomorphic to
a normal subgroup of
$SL_2(\mathbb{Z}/p\mathbb{Z})$, i.e.
$N$ is trivial, equal to $Z$, or
isomorphic to
$SL_2(\mathbb{Z}/p\mathbb{Z})$. In
the latter case we would have
$G=K_1\times N$ and it is easy to
check that this does not happen. Thus
$N$ is trivial or equal to $Z$.

*Assume next that $N\cap K_1$ is
non-trivial. In particular $N\cap K_1$ is
a non-trivial normal subgroup of
$K_1$. We use the following easy
fact: A non-trivial normal subgroup
of a $p$-group intersects the center
of that $p$-group non-trivially. Thus
$N\cap K_{n-1}$ is non-trivial and is
a normal subgroup of $G$. But, since
$SL_2(\mathbb{Z}/p\mathbb{Z})$ acts
irreducibly on the module $N\cap
   K_{n-1}$, $N$ must contain $K_{n-1}$.
If $N\cap K_1 = K_{n-1}$, then there
are three possibilities for $N$, namely $N=
   K_{n-1}$, $N=K_{n-1} \times Z$ or $N=K_{n-1}.
   SL_2(\mathbb{Z}/p\mathbb{Z})$. If $n=2$, the last possibility corresponds to $G$ and we are done. If $n>2$, then the
last possibility is impossible just
as before.

*Now the proof is completed by
observing that $G/K_{n-1} \cong
   SL_2(\mathbb{Z}/p^{n-1}\mathbb{Z})$ and
appealing to induction.


Final remark: I've read some of Bourgain & Gamburd's work dealing with $SL_2$. They tend to (implicitly) consider the center as a trivial normal subgroup as their work deals with asymptotics on $p$ which are unaffected by $Z$. This explains the apparent inaccuracy of their assertion that the $K_i$ are all of the normal subgroups of $SL_2(\mathbb{Z}/p^n\mathbb{Z})$.
