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

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 $SL_2(\mathbb{Z}/n)$?

• 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. – Derek Holt Jan 14 '17 at 4:55
• @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? – stupid_question_bot Jan 15 '17 at 2:04
• I checked it computationally for $m \le 5$. ${\rm SL}(2,Z/32)$ has $37$ normal subgroups. – Derek Holt Jan 15 '17 at 10:24
• @DerekHolt Ah, excellent. I also checked it for $m\le 6$, so it seems likely. – stupid_question_bot Jan 16 '17 at 2:42