Let's fix an algebraically closed field $k$ of arbitrary characteristic and a connected nonsingular affine algebraic $k$-group $G$. Under what conditions can I assume that a connected nonsingular affine algebraic subgroup $H \subseteq G$ is a complete intersection in $G$? Or are such subgroups always complete intersections? (I would imagine not, but I don't know any counterexamples.) If it helps I'm happy to assume that $G$ is semisimple.


EDIT: I should probably say "smooth k-group scheme" here instead of "nonsingular."



Smooth affine algebraic subgroups as complete intersections
This answer is just to record that Will Sawin's example works: The Borel subgroup of $PGL\_2$ is not a complete intersection in $PGL\_2$. Recall that the coordinate ring of $GL\_2$ is $k[w,x,y,z,\Delta^{-1}]$ where $\Delta=wz-xy$. We are thinking of $w$, $x$, $y$, $z$ as entries of the matrix $\left( \begin{smallmatrix} w & x \\ y & z \end{smallmatrix} \right)$. The coordinate ring of $PGL\_2$ is the subring of homogenous elements, where $w$, $x$, $y$ and $z$ are graded in degree $1$ and hence $\Delta^{-1}$ has degree $-2$. 


The Borel is given by the equations $wy/\Delta=xy/\Delta=y^2/\Delta=yz/\Delta=0$. Conceptually, the defining equation is $y=0$, but the element $y$ isn't in the coordinate ring of $PGL\_2$. Since the Borel is a hypersurface, if it were a complete intersection, it would be defined by the vanishing of a single equation, say $g(w,x,y,z)/\Delta^k$, where $g$ is a homogeneous polynomial of degree $2k$. We may also assume that $\Delta$ does not divide $g$. Since $g$ has even degree, it must either vanish to even order along $y=0$, or else vanish along some other hypersurface besides $y=0$. Either way, $g/\Delta^k$ does not define the ideal of the Borel.


