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


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.

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    $\begingroup$ I'm guessing you really mean "unable to prove it" $\endgroup$ – Geoff Robinson Aug 8 '12 at 8:46
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    $\begingroup$ You (at least) also have all subgroups of the group of scalar matrices. $\endgroup$ – YCor Aug 8 '12 at 11:32
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    $\begingroup$ ... as well as some inverse images of groups of scalar matrices (e.g. the normal subgroup consisting of matrices whose reduction modulo $p^r$ is scalar). Note that for $k=p=2$ and $n>2$, the scalar matrices are not only plus and minus identity, since $x^2=1$ has 4 solutions in $Z/2^nZ$. If your question is whether every normal subgroup is obtained this way, you should edit accordingly. $\endgroup$ – YCor Aug 8 '12 at 15:30
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    $\begingroup$ But my first comment as well as the first part of my comment applies for all $p$: you have inverse images of scalar matrices mod $p^r$. $\endgroup$ – YCor Aug 8 '12 at 23:40
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    $\begingroup$ Have a look at Theorem 7.5 in Bass, Milnor, Serre, Solution of the congruence subgroup problem for $SL_n$, (n≥3) and $Sp_{2n}$, (n≥2). $\endgroup$ – Wilberd van der Kallen Aug 9 '12 at 9:19

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 $SL_2(\mathbb{Z}/p^n\mathbb{Z})$.

  • $\begingroup$ Dear Nick: Many thanks for your answer. Do you know any reference which answers this question for general rings? I am in particular interested in the structure of normal subgroups of $\mathrm{SL}_k(\mathcal{O}/\mathfrak{p}^n)$, $k\geq 3$, where $\mathcal{O}$ is the ring of integer of a non-archimedean local fields of characteristic $0$, with maximal ideal $\mathfrak{p}$. Assume that the characteristic of $\mathcal{O}/\mathfrak{p}$ is not $2$. $\endgroup$ – M.B Sep 17 '14 at 14:06
  • $\begingroup$ @M.B Sorry, no idea. I don't know anything about the fields you describe... $\endgroup$ – Nick Gill Sep 17 '14 at 22:46

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