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. 



Normal subgroup of classical groups
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 $SL\_2(\mathbb{Z}/p^n\mathbb{Z})$.


