is "being reductive" extension-closed? Suppose we have a short exact sequence of linear algebraic groups over a field of characteristic zero $$1 \to N \to G \to G/N \to 1$$ with $N$ and $G/N$ reductive (that is connected with trivial unipotent radical over the algebraic closure of the base field). > Is $G$ also reductive?

The answer is yes:

Let $U \trianglelefteq G$ be connected unipotent. Then the image of $U$ in $G/N$ is also connected unipotent normal, so it is trivial because $G/N$ is reductive. In other words $U \trianglelefteq N$. But $N$ is also reductive so $U$ is trivial.