A group $G$ is said to be linear if there exists a field $k$, an integer $n$ and an injective homomorphism $\varphi: G \to \text{GL}\_n(k).$ 
Given a short exact sequence 
$1 \to K \to G \to Q \to 1$ of groups where $K$ and $Q$ are linear (over the same field), is it true that $G$ is linear too?


Background: [Arithmetic groups](http://en.wikipedia.org/wiki/Arithmetic_group) are by definition commensurable with a certain linear group, so they are finite extensions of a linear group, and finite groups clearly are linear (over any field).



Are extensions of linear groups linear?
The universal central extension $\widetilde{\text{Sp}\_{2n}}\mathbb{Z}$ is the preimage of $\text{Sp}\_{2n}\mathbb{Z}$ in the universal cover of $\text{Sp}\_{2n}\mathbb{R}$, and fits into the sequence


$$1\to \mathbb{Z}\to \widetilde{\text{Sp}\_{2n}}\mathbb{Z}\to \text{Sp}\_{2n}\mathbb{Z}\to 1.$$


Deligne proved that $\widetilde{\text{Sp}\_{2n}}\mathbb{Z}$
is not residually finite; the intersection of all finite-index subgroups of is $2\mathbb{Z}<\widetilde{\text{Sp}\_{2n}}\mathbb{Z}$. In particular, this implies that $\widetilde{\text{Sp}\_{2n}}\mathbb{Z}$ is not linear. But certainly $\mathbb{Z}$ and $\text{Sp}\_{2n}\mathbb{Z}$ are. If you want an arithmetic group, you can take the corresponding $\mathbb{Z}/k\mathbb{Z}$-extension of $\text{Sp}\_{2n}\mathbb{Z}$, which will not be linear as long as $k\neq 2$.


I learned the proof of this theorem from Dave Witte Morris, who has written up his fairly-accessible notes as "[A lattice with no torsion-free subgroup of finite index](http://people.uleth.ca/~dave.morris/talks.shtml) (after P. Deligne)" ([PDF](http://people.uleth.ca/~dave.morris/talks/deligne-torsion.pdf) link).


