The starting point of this question is the (presumably) well-known theorem (the proof I know is from *Abelian $\ell$-adic representations and elliptic curves* from J-P.Serre in which it is a lemma for $n=2$ and an exercise for $n>2$; which suggests that the result was already classical in the 60s).


**Theorem I:** If $G$ is a closed subgroup of $\operatorname{SL}\_{n}(\mathbb Z\_{p})$ which surjects on $\operatorname{SL}\_{n}(\mathbb F\_{p})$ and if $p≥5$, then $G=\operatorname{SL}\_{n}(\mathbb Z\_{p})$. 


The theorem is optimal with respect to all hypotheses in the sense that there exists proper subgroups of $\operatorname{SL}\_{2}(\mathbb Z\_{p})$ mapping onto $\operatorname{SL}\_{2}(\mathbb F\_{p})$ when $p=2,3$ (a fact that played a role in the original proof of the modularity of semi-stable elliptic curves by A.Wiles, if I am not mistaken) and in the sense that for all $n≥2$ and all prime $p$, there exist discrete valuation rings $A$ of mixed characteristic $(0,p)$ such that $\operatorname{SL}\_{n}(A)$ contains proper subgroups mapping onto $\operatorname{SL}\_{n}(A/\mathfrak m)$ (just take a completely ramified $A$ over $\mathbb Z\_{p}$ and consider $\operatorname{SL}\_{2}(\mathbb Z\_{p})$ inside $\operatorname{SL}\_{2}(A)$).


A slightly less well-known fact is that the theorem admits the following generalization, due to N.Boston.


**Theorem II:** Let $A$ be a complete local noetherian ring with finite residual characteristic $p\neq 2$. If $G$ is a closed subgroup of $\operatorname{SL}\_{n}(A)$ which surjects on $\operatorname{SL}\_{n}(A/\mathfrak m^2)$, then $G=\operatorname{SL}\_{n}(A)$. 


This is in the appendix of *On p-adic analytic families of Galois representations*. 
Compositio Math. 59 (1986), no. 2, 231–264 by B.Mazur and A.Wiles and again, this theorem is optimal in the sense that there exists a proper subgroup of $\operatorname{SL}\_{2}(\mathbb Z\_{2})$ surjecting on $\operatorname{SL}\_{2}(\mathbb Z/4\mathbb Z)$. 


Now my actual question. Let $A$ be a complete local noetherian ring (UPDATE: domain, actually) of mixed characteristic $(0,p)$ with $p≠2$ UPDATE: which one can assume to be a discrete valuation ring if necessary. Suppose $G$ is a closed subgroup of $\operatorname{SL}\_{n}(A)$ which surjects on $\operatorname{SL}\_{n}(A/\mathfrak m)$. 



> 
> Among the pre-images in $G$ of non-identity unipotents elements in $\operatorname{SL}\_{n}(A/\mathfrak m)$, is it true that there exists a unipotent element (that is to say an element $\sigma$ such that $\sigma-1$ is nilpotent)?
> 
> 
> 


Granted theorem II, this is obviously true if $G$ maps onto $\operatorname{SL}\_{n}(A/\mathfrak m^2)$. Without this hypothesis, it looks dubious to me but nevertheless, the obvious counterexamples to theorem I coming from ramified rings do not provide counterexamples to this claim and I confess that I don't quite know how to construct other counter-examples. Taking this into account, another more general point of view on the question would be the following.



> 
> What are the subgroups of $\operatorname{SL}\_{n}(A)$ which do not map onto $\operatorname{SL}\_{n}(A/\mathfrak m^2)$ but which do map onto $\operatorname{SL}\_{n}(A/\mathfrak m)$? 
> 
> 
> 


Perhaps group cohomology of $\operatorname{SL}\_{n}$ would help then. Any positive result, even in the case $n=2$ and $p>3$ would already be of interest to me. 


UPDATE: Jim Humphreys asks in comments what I mean by a unipotent element in a matrix group. I meant an element $\sigma$ such that $\sigma-1$ is nilpotent, but I realize now that this might not be so great a definition when the ring of coefficients is not a domain, or at least reduced. So perhaps it is better to take $A$ a domain in the above, or even a discrete valuation ring. Also, even though Tim Dokchister answered the first question in the negative, I would be interested to know if someone has something to say for infinite closed subgroup $G$.



Pre-images of unipotent elements in $\operatorname{SL}_{n}(A)$
For the first question, at least for $p=5, n=2$ there is a counterexample.:


The group $\text{SL}\_2({\mathbb F}\_5)$ has a 2-dimensional symplectic representation with character in ${\mathbb Q}(\sqrt 5)$ and Schur index 2, so it is realizable over $A={\mathbb Z}\_5(\zeta\_5)$. So $\text{SL}\_2({\mathbb F}\_5)$ injects in $\text{SL}\_2(A)$ and reduces onto $\text{SL}\_2(A/{\mathfrak m})$. But it is just a finite group, so it has no non-identity unipotent elements.


I don't know whether it generalizes to higher $p$ and $n$ though, and whether there are examples with `interesting' infinite closed subgroups of $\text{SL}\_2(A)$.


