$\DeclareMathOperator\SL{SL}$ $\DeclareMathOperator\GL{GL}$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 $1 \neq -1$.



Does the map $\mathrm{SL}(n,\mathbb{Z}/p^2) \rightarrow \mathrm{SL}(n,\mathbb{Z}/p)$ split?
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**](https://doi.org/10.1016/0021-8693(74)90099-4), J. Algebra 29, 255-302 (1974). [ZBL0277.20071](https://zbmath.org/?q=an:0277.20071).


