Question on subgroups of reductive groups A linear algebraic group $G$ over some field $k$, which I assume being of characteristic 0, is _reductive_ if $R_u(G^0_{\overline{k}})$ is trivial, where $R_u$ denotes the unipotent radical, $G^0$ is the connected component of the identity, and $G_{\overline{k}}= G \times_k \overline{k}$. My question: > Is any subgroup of a reductive group reductive? My thoughts: let $H \leq G$. Then $H^0 \leq G^0$. If $R_u(H^0_{\overline{k}})\leq R_u(G^0_{\overline{k}})$, then I'm done, but I don't know if this holds.

No. Any linear algebraic group is (by definition) a subgroup of $GL_n$; $GL_n$ is reductive, but there exist linear algebraic groups that are not reductive. Ergo, there are non-reductive subgroups of reductive groups.

For an explicit example try the subgroup of upper-triangular unipotent matrices $H=\begin{pmatrix} 1 & x \\\ 0 & 1 \end{pmatrix}$ in $G=SL_2$.