$\DeclareMathOperator\SL{SL}\DeclareMathOperator\GL{GL}$Is every normal subgroup of $\SL_2(\mathbb{Z}/n)$ also normal inside $\GL_2(\mathbb{Z}/n)$?

Of course, it suffices to ask the question when $n = p^m$ a prime power. In Classification of the Normal Congruence Subgroups of the Modular Group, McQuillan describes the normal subgroups of $\SL_2(\mathbb{Z}/p^m)$. When $p >3$, the only nontrivial normal subgroup is $\{\pm I\}$ (together with the kernels of maps to smaller prime powers). When $p = 3$, there is one additional possibility, of index 3, which one can check to be normal in $\GL_2(\mathbb{Z}/3^m)$.

When $p = 2$, on page 292 he describes the various normal subgroups of $\SL_2(\mathbb{Z}/2^m)$. However, amongst these possibilities, he writes that for $m\ge 4$, the matrix: $$A := \begin{bmatrix}1+2^{m-2} & 2^{m-1} \\ 2^{m-1} & 1-2^{m-1}\end{bmatrix}$$ generates a normal subgroup of $\SL_2(\mathbb{Z}/2^m)$, which seems to be incorrect as checked in GAP...which makes me a bit wary of his results.

In any case, for my purposes, I don't need a full classification, and I suspect my question probably has a simpler answer anyway.

EDIT: Actually I suppose a much weaker question is relevant for my purposes - Is it true that every normal subgroup of $\SL_2(\mathbb{Z}/n)$ is the intersection of a normal subgroup of $\GL_2(\mathbb{Z}/n)$ with $\SL_2(\mathbb{Z}/n)$? (As YCor points out in the comments, this is trivially equivalent to the original question. Living up to my name...)

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    $\begingroup$ You don't mean the only nontrivial normal subgroup is $\{ \pm 1 \}$, because the kernels of the maps onto smaller ${\rm SL}_2(Z/p^k)$ are also normal subgroups. But these and combinations with $\{ \pm 1\}$ are the only normal subgroups. The situation is similar with $p=3$. With $p=2$ there are more normal subgroups (essentiallly because the three dimensional module is reducible when $p=2$) but they all appear to be normal in ${\rm GL}$, so the answer seems to be yes. $\endgroup$
    – Derek Holt
    Commented Jan 14, 2017 at 4:55
  • $\begingroup$ @DerekHolt Yes, thanks I've edited the question accordingly. By "they all appear to be normal", do you mean that you checked some values of $m$ computationally, or do you have a reference? $\endgroup$ Commented Jan 15, 2017 at 2:04
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    $\begingroup$ I checked it computationally for $m \le 5$. ${\rm SL}(2,Z/32)$ has $37$ normal subgroups. $\endgroup$
    – Derek Holt
    Commented Jan 15, 2017 at 10:24
  • $\begingroup$ @DerekHolt Ah, excellent. I also checked it for $m\le 6$, so it seems likely. $\endgroup$ Commented Jan 16, 2017 at 2:42
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    $\begingroup$ The "much weaker question" is actually equivalent to the original question. In general, you have a group $G$ and a normal subgroup $S$. The first question is whether every normal subgroup of $S$ is normal in $G$. The second is whether every normal subgroup of $S$ is intersection of a normal subgroup of $G$ with $S$. This is equivalent, since this intersection is then normal in $G$. $\endgroup$
    – YCor
    Commented Oct 6, 2023 at 6:25

1 Answer 1


The answer to the question in the title is yes.

As you already recognized, this can be solved by using the paper by McQuillan. You state that you distrust this analysis because of a supposed error in it.

However, the proof given there seems reliable to me after reading it -- although there is a typo in McQuillan's definition of the matrix $A$ on page 292 (but note that on page 286, it appears correctly): in the lower right entry the exponent should be $m-2$ not $m-1$. Assuming that you entered the "bad" version into GAP, that would explain what you observed. But this is merely a typo, the analysis given in the rest of the paper appears solid.

(NOTE: in the initial version of this answer I claimed that the paper gave the correct version of the matrix $A$, as I only was aware of the correct one on page 286, and did not notice that $A$ was given incorrectly on page 292. So this is really an error in the paper. I apologize for giving the incorrect impression that the @stupid_question_bot made a mistake here -- they did not!)

But using the correct matrix for $A$, GAP indeed confirms that it generates a normal subgroup in $SL_2(\mathbb{Z}/2^m)$ and also in $GL_2(\mathbb{Z}/2^m)$, at least for a finite set of values of $m$ (here I checked $4 \leq m \leq 70$ but the upper bound was arbitrary):

gap> testit:=function(m)
>   local R, A, G, H;
>   R:=Integers mod 2^m;
>   A := One(R) * [[ 1+2^(m-2), 2^(m-1)], [2^(m-1), 1-2^(m-2)]];
>   G := GL(2,R);
>   H := Group(A);
>   # this is inefficient in GAP right now:
>   #   return IsNormal(G,H);
>   # ... so we do a naive but in this case efficient test instead
>   return ForAll(GeneratorsOfGroup(G), g -> A^g in H);
> end;;
gap> ForAll([4..70], testit);

But of course we can also just verify that this is really a normal subgroup by hand, by following the proof McQuillan gives, or by taking any generating set of $GL_2(\mathbb{Z}/2^m)$ and verifying that it indeed normalizes $A$. One such generating set is $$ T=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix},\quad S=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix},\quad M=\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix},\quad D=\begin{pmatrix} 5 & 0 \\ 0 & 1 \end{pmatrix}. $$ Note that the group generated by $A$ has order 4, and is distinct from that generated by $-A$, but both are normal, as one can now easily verify that $A^S=A^3=A^T$ (actually McQuillan already states that), and $A^M=A=A^D$.

  • $\begingroup$ $A^B$ means $B^{-1}AB$? $\endgroup$ Commented Oct 6, 2023 at 2:37
  • $\begingroup$ Yes exactly, conjugation $\endgroup$
    – Max Horn
    Commented Oct 6, 2023 at 5:10
  • $\begingroup$ The answer to the question in the title and in the first line of the question. I'll clarify. $\endgroup$
    – Max Horn
    Commented Oct 6, 2023 at 7:56
  • $\begingroup$ Bizarre! The published version in American Journal of Math seems to agree with what I wrote in the OP. Which version are you looking at? $\endgroup$ Commented Oct 8, 2023 at 6:16
  • $\begingroup$ I followed the link you gave, so we should be looking at the same PDF. But indeed, on page 292, the wrong matrix (the same as in your question) is given, which I did not see before -- I was looking at the correct (but projective) version given on page 286, and did not notice that the second version had the mistake. My apologies for the mischaracterization, I shall edit my answer accordingly right away. $\endgroup$
    – Max Horn
    Commented Oct 8, 2023 at 7:41

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