Does the map $\mathrm{SL}(n,\mathbb{Z}/p^2) \rightarrow \mathrm{SL}(n,\mathbb{Z}/p)$ split?

Question

$\DeclareMathOperator\SL{SL}$ $\DeclareMathOperator\GL{GL}$The question is the one in the title: for a prime $p$, does the obvious surjection $\pi\colon \SL(n,\mathbb{Z}/p^2) \rightarrow \SL(n,\mathbb{Z}/p)$ split?

Actually, I know that the answer is "no" for $p \geq 5$. The point is that in that case, there is no way to choose a lift $\tilde{E} \in \SL(n,\mathbb{Z}/p^2)$ of an elementary matrix $E \in \SL(n,\mathbb{Z}/p)$ such that $\tilde{E}$ has order $p$. Unfortunately, this is possible for $p=2$ and $p=3$, so those are my primary interests (but I would also be interested in references for the large primes case, since I am sure I am not the first person to notice the above obstruction).

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EDIT 3: Excluding the most recent answer (which seems to slightly contradict the others, and which I am going through right now), we now have dealt with everything but $\SL(n,\mathbb{Z}/4) \rightarrow \SL(n,\mathbb{Z}/2)$ for $n \geq 4$. I did a brute-force computation with Mathematica, and assuming I didn't make any errors it shows that there indeed does not exist a lift of the upper triangular matrices (and hence $\SL(n,\mathbb{Z}/2)$ itself) for $n=4$, even if we allow determinant $-1$.

EDIT 2: The edit below is slightly wrong. What it proves is that the map $\GL(2,\mathbb{Z}/4) \rightarrow \GL(2,\mathbb{Z}/2)$ splits. As people noted below, this doesn't seem to hold for $\SL$.

EDIT: Something which I forgot to put in the original version of the question is that this does split for $p=2$ when $n=2$. Assuming I have done the calculations correctly, a splitting homomorphism $\sigma\colon \SL(2,\mathbb{Z}/2) \rightarrow \SL(2,\mathbb{Z}/4)$ can be defined in the following way. On the elementary matrix generators for $\SL(2,\mathbb{Z}/2)$, we define $\sigma$ as follows:

$$\sigma\left(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}\right) = \left(\begin{matrix} 1 & 1 \\ 0 & -1 \end{matrix}\right)$$ and $$\sigma\left(\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}\right) = \left(\begin{matrix} 1 & 0 \\ 1 & -1 \end{matrix}\right).$$

I don't have any deep explanation as to why this works: all I did was perform some linear algebra to find matrices in $\SL(2,\mathbb{Z}/4)$ satisfying the relations in $\SL(2,\mathbb{Z}/2)$ between elementary matrices. There is actually quite a bit of flexibility in this. More generally, for constants $a,b,c \in 2 \mathbb{Z}/4\mathbb{Z}$ you can also take $\sigma$ as follows:

$$\sigma\left(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}\right) = \left(\begin{matrix} 1+a & 1+b \\ 0 & -1+a \end{matrix}\right)$$ and $$\sigma\left(\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}\right) = \left(\begin{matrix} 1+c & 0 \\ 1+b & -1+c \end{matrix}\right).$$

I haven't done the calculation to see if you can also split it for $p=3$ when $n=2$ — I was able to do the above by hand, but I would need to install Mathematica or something to do all the matrix multiplications in a setting where $1 \neq -1$.

Answer 1

The map splits if and only if $(n,p)$ is either $(2,3)$ or $(3,2)$. As far as I know, this is a theorem of C.-H. Sah. See Theorem 7 in

Sah, Chih-Han, Cohomology of split group extensions, J. Algebra 29, 255-302 (1974). ZBL0277.20071.

Answer 2

If $p=2$ then it is easy to check that there is no inverse image of $\left(\begin{smallmatrix} 1&1\\0&1\end{smallmatrix}\right)$ of order two, so there is no splitting. If $p=3$ then it actually splits all the way to the $3$-adics, as $SL(2,3)$ is a $3$-adic reflection group lifting the natural representation (the binary tetrahedral group, number $4$ in the Shephard-Todd list).

Edit: actually I don't think the $3$-adics are right. You need cube roots of unity, so the lift is to $SL\left(2,{\mathbb Z}\left[\frac{1+i\sqrt {3}}{2}\right]\right)$. But the cube roots of unity exist modulo $9$, which answers the original question.

Edit 2: The $3$-adics do work! See David E. Speyer's comments below.

Answer 3

• For $n=2$: (as already said in comments) it splits iff $p=3$. Indeed by Gaschütz, this reduces to whether the matrix $e_{12}(1)$ lifts to an element of order $p$. For $p\ge 5$ it doesn't lift at all to an element of $\mathrm{GL}_2(\mathbf{Z}/4\mathbf{Z})$. For $p=2$ it does, but not to a determinant 1 one. For $p=3$, it lifts to the determinant 1 matrix of order 3 $\begin{pmatrix}-2 & 1\\-3 & 1\end{pmatrix}$.

• For $n\ge 3$ and $p=3$: it doesn't split. It it enough in this case to show that the elementary abelian 3-group of order 9 $\langle e_{12},e_{13}\rangle$ doesn't split to $\mathrm{GL}_n(\mathbf{Z}/p^2\mathbf{Z})$. That is, $e_{12}$ and $e_{13}$ can't lift to commuting elements of order 9:

First, a formula in an associative unital ring with $9=0$ is $(B+3A)^3=B^3+3(B^2A+BAB+AB^2)$. If $B=1+N$ with $N^2=0$, this writes $(1+N+3A)^3=1+3(N+NAN)$.

Now in the matrix algebra, we have $E_{ij}AE_{ij}=A_{ji}E_{ij}$. So taking $N=E_{ij}$, the condition $(1+N+3A)^3=0$ means $A_{ji}=-1$ mod 3. That is, mod 3, $A=-E_{ji}+A'$ with $A'_{ji}=0$.

Now lift $e_{12}$ and $e_{13}$ to elements of order 3 mod 9: these are $u=I+E_{12}-3E_{21}+3U$ and $v=I+E_{13}-3E_{31}+3V$, with $U_{21}=V_{31}=0$ mod 3.

Now express the condition that $u,v$ commute, i.e. $[u,v]:=uv-vu=0$: $$0=[E_{12}-3E_{21}+3U,E_{13}-3E_{31}+3V]$$ $$=3\big(-[E_{12},E_{31}]-[E_{21},E_{13}]+[E_{12},V]-[U,E_{13}]\big),$$ so, mod 3 $$0=E_{32}-E_{23}+E_{12}V-VE_{12}+UE_{13}-E_{13}U.$$ Taking the $(3,2)$ entry, we get, modulo 3 $$0=1-0+0-(VE_{12})_{32}+0+0-0=1-V_{31}=1,$$ contradiction.

Edit:

• for $n=3$, $p=2$ it lifts (see D. Speyer's answer

• for $n=4$, $p=2$: it doesn't lift:

Namely, the Jordan matrix $I+J=\begin{pmatrix}1 & 1 & 0 & 0\\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 1\end{pmatrix}$, of order 4 modulo 2, doesn't lift to a determinant 1 matrix of order 4.

This is just a computation: let a lift (mod 4) be $u=\begin{pmatrix}1+2a & 1+2b & 2c & 2d\\ 2e & 1+2f & 1+2g & 2h \\ 2i & 2j & 1+2k & 1+2l\\ m & n & o & 1+2p\end{pmatrix}$.

Then computation yields $$\det(u)=1+2(a+f+k+p)+2(e+j+o)+2(i+n)+2m;$$ also we have the formula (just using we are in an associative unital ring with $4=0$) $(1+J+2A)^4=1+J^4+2(J^2+J^3A+J^2AJ+JAJ^2+AJ^3)$.

Here, for $A=\begin{pmatrix}a & b & c & d\\ e & f & g & h \\ i & j & k & l\\ m & n & o & p\end{pmatrix}$ and $J$ as above (so that $J^4=0$, $u=I+J+2A$) we have $$J^3A+J^2AJ+JAJ^2+AJ^3=\begin{pmatrix}m & i+n & e+j+o & a+f+k+p \\ 0 & m & i+n & e+j+o \\ 0 & 0 & m & i+n \\ 0 & 0 & 0 & m\end{pmatrix}$$ the condition that it is equal mod 2 to $J^2$ yields, mod 2, $m=i+n=a+f+k+p=0$, $e+j+o=1$.

But this yields $\det(u)=-1$.

• I haven't yet checked whether such a matrix fails to lift to a matrix of the same order for $n\ge 5$.

Answer 4

According to Theorem 2.5 in https://doi.org/10.1090/mcom/3236 , the only cases when the map $SL(n,p^2)\to SL(n,p)$ splits are $$ SL(2,Z_9), SL(3, Z_4) $$ ($SL(2,Z_4)$, $SL(4,Z_4)$ are listed in the theorem as having a supplement, but do not have a complement.)

One can find the complements in each case easily with GAP's function ComplementClassesRepresentatives.

Answer 5

$\def\SL{\text{SL}}$For $n=3$, $p=2$, there is a splitting. $\SL_3(\mathbb{F}_2)$ is the simple group of order $168$. It has a $3$ dimensional representation with coefficients in $\mathbb{Z}\left[\tfrac{1+\sqrt{-7}}{2} \right]$. (See Elkies, Section 1.1.) Since $\sqrt{-7}$ exists in the $2$-adics, this representation can be defined over $\mathbb{Z}_2$ (and, in particular, over $\mathbb{Z}/2^k \mathbb{Z}$ for any $k$.)

YCor and Dave Benson have already addressed all of the $p=3$ cases, so what remains is $p=2$, $n \geq 4$.

$\text{SL}_4(\mathbb{F}_2)$ does not lift all the way to the $2$-adics: It is isomorphic to the alternating group $A_8$ and, in characteristic $0$, the smallest nontrivial representation is $7$-dimensional. But I don't know how to figure out whether it lifts to $\SL_4(\mathbb{Z}/4 \mathbb{Z})$.

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For the record, I'll write up the computation that rules out $p \geq 5$. Let $N$ be a matrix with a single $1$ in position $(i, i+1)$. We are looking for solutions to $$(\text{Id}+N + p X)^p \equiv \text{Id} \bmod p^2.$$ Since $p^2$ and $N^2$ are $0$, the left hand side is just $$\text{Id} + p N + p \binom{p}{2} ( NX+XN ) + p \binom{p}{3} NXN.$$ If $p \geq 5$, then $\binom{p}{2} \equiv \binom{p}{3} \equiv 0 \bmod p$, and this just collapses to $\text{Id}+pN$, giving no solutions.