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 are by definition commensurable with a certain linear group, so they are finite extensions of linear group, and finite groups clearly are linear (over any field).

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 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).

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, is it true that $G$ is linear too?

Background: Arithmetic groups are by definition commensurable with a certain linear group, so they are finite extensions of linear group, and finite groups clearly are linear.

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 are by definition commensurable with a certain linear group, so they are finite extensions of linear group, and finite groups clearly are linear (over any field).