Does there exist a section of $GL_n(\mathbb{Z}/9\mathbb{Z}) \rightarrow GL_n(\mathbb{Z}/3\mathbb{Z})$? There is the reduction map $r : GL_n(\mathbb{Z}/9\mathbb{Z}) \rightarrow GL_n(\mathbb{Z}/3\mathbb{Z})$. When does there exist a group homomorphism $i : GL_n(\mathbb{Z}/3\mathbb{Z}) \rightarrow GL_n(\mathbb{Z}/9\mathbb{Z})$ such that $r \circ i = id$ ?
After testing on Python I think that there exists such section for $n = 2$ but not for $n \geq 3$, but I don't manage to prove it.
I tried to use the fact that the group $GL_n(\mathbb{Z}/3\mathbb{Z})$ is generated by two matrices : an elementary matrix $E$ and a permutation matrix $P$ that acts as a $n$-cycle on the basis (modulo a sign, depending if $n$ is odd or even). I wanted to prove that there doesn't exist any $(i(E),i(P))$ in $GL_n(\mathbb{Z}/9\mathbb{Z})$ which respect the group law. However according to Python for $n=3$, there exists for instance $(i(E),i(P))$ which preserve at least the order of $E$, $P$, $PE$, $P^2E$, so it seems that an argument using this method would be quite complicated...
I managed to prove that there doesn't exist a section of $GL_n(\mathbb{Z}/p^2\mathbb{Z}) \rightarrow GL_n(\mathbb{Z}/p\mathbb{Z})$ for any $n \geq 2$ if $p \geq 5$, but the proof doesn't work if $p =3$.
 A: I’m not entirely sure, but this might work.
Let $R=\mathbb{Z}/9\mathbb{Z}$ and $k=R/3R=\mathbb{F}_3$ the residue field.
Claim: if a section of $GL_n(R) \rightarrow GL_n(k)$ exists, then we can assume that it maps diagonal matrices (with only $1$ or $-1$ on the diagonal) to themselves.
Proof: this is the usual codiagonalization argument over a field, with a few twists: such diagonal matrices are of square $I_n$, so their “naive” eigenspaces are in direct sum and sum to the full space, $R$ is local so such eigenspaces are free.
Usually, codiagonalization doesn’t give us anything about the kind of basis we get, but here, we know that it has to be the canonical basis mod $3$, which concludes.
Unless explicitly mentioned otherwise, we will always consider sections mapping diagonal matrices to themselves.
Let $A_i$ be the diagonal matrice with only ones on the diagonal, except the $(i,i)$-th coefficient which is $-1$.
Let $n \geq 3$ and assume we have a section $s$. For $1 \leq i \leq n$, we have a subgroup $\beta_i(R)=GL_{n-1}(R) \times R^{\times}$ of $GL_n(R)$ made with “block matrices”, ie that are zero at indices $(i,j)$ and $(j,i)$ for $j \neq i$. We can make a similar definition for $GL_n(k)$ (and we call the subgroup $\beta_i(k)$).
(It’s a but easier to see for $i=1$ and $i=n$, you can restrict to this case first if you want).
Now, I claim that on $R$ or $k$, $\beta_i$ is constituted with the matrices that commute to $A_i$ (it’s a simple enough computation). Thus, $s(\beta_i(k)) \subset \beta_i(R)$. By taking the entries in $(\{ 1;\ldots;n\}\backslash \{i\})^2$, we recover a section $s_i: GL_{n-1}(k) \rightarrow GL_{n-1}(R)$.
Therefore, it is enough to show that there is no section for $n=3$ – we now assume $n=3$.
Now, if $(M,\alpha) \in \beta_i(k)$, write $s(M,\alpha)=(\sigma_i(M,\alpha),\lambda_i(M,\alpha)) \in \beta_i(R)$, and $\sigma_i: \beta_i(k) \rightarrow GL_2(R)$, $\lambda_i: \beta_i(k) \rightarrow R^{\times}$ are group homomorphisms.
But $\sigma_i(I_{n-1},-1)=I_{n-1}$ and $\lambda_i(I_{n-1},-1)=-1$ (this corresponds to the fact that $s(A_i)=A_i$), so that $\sigma_i(M,\alpha)=\sigma_i(M,1)$, $\lambda_i(M,\alpha)=\alpha\lambda_i(M,1)$. Because $s$ is a section of the reduction in $k$, it follows that $\lambda_i(M,1) \in R^{\times}[3]$.
But in $GL_2(\mathbb{F}_3)$, the unipotent matrix is generated by commutators; moreover, the restriction of $\lambda(-,1)$ to the $2$-Sylow (of order $16$) must be trivial (since $16$ and $3$ are coprime), thus $\lambda=1$, and $s((M,\alpha) \in \beta_i(k))=(\sigma_i(M),\alpha) \in \beta_i(R)$.
$SL_3(\mathbb{F}_3)$ is perfect, so $\det{s}$ must factor through $\det$; therefore (see the images of the $A_i$), $s$ preserves $\det$ thus so must every $\sigma_i$.
Moreover, let $\sigma \in S_3$ be a permutation. For a given matrix $M$ (over $R$ or $k$), $M$ has “support in $\sigma$” (ie $M_{ij}=0$ whenever $i \neq \sigma(j)$) iff for all $j$, $MA_j=A_{\sigma(j)}M$. It follows that $s$ maps matrices with support in $\sigma$ to matrices with support in $\sigma$.
Let $\begin{bmatrix} 0&0&\alpha\\\beta&0&0\\0&\gamma&0\end{bmatrix}$ be the image of $c=\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}$ under $s$, then $\alpha,\beta,\gamma \in R$ are congruent to $1$ mod $3$ and have product one.
Then we can find $u,v,w \in R^{\times}$ congruent to $1$ mod $3$ such that $w/u=\alpha$, $u/v=\beta$, $v/w=\gamma$, and thus $\mathrm{diag}(u,w,v)s(c)\mathrm{diag}(u^{-1},w^{-1},v^{-1})=c$. Up to conjugating (by a diagonal matrix congruent to $I_3$ mod $3$, so everything else still works), we can thus assume $s(c)=c$. From this, we deduce (working out what happens for “transposition matrices”) that $s$ (and therefore all $\sigma_i$) preserves all permutation matrices too.
A computer search shows that this gives a single possibility for $\sigma_i$, which, in particular, maps $\begin{bmatrix}1&1\\0&1\end{bmatrix}$ to $\begin{bmatrix} 4&7\\6&4\end{bmatrix}$. Now consider the product $\begin{bmatrix}1&1&0\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}$. The result has multiplicative order $3$ (in $GL_3(k)$) but (unless I made a mistake), its prescribed image by $s$ (both factors are block matrices) is not the identity matrix when cubed. We get our contradiction.
