Let $G$ be a connected algebraic group over a field of characteristic $p \geq 0$ and let $H < G$ be a connected closed subgroup.
If the lie algebra $\mathfrak{h}$ of $H$ is an ideal of the Lie algebra $\mathfrak{g}$ of $G$, is $\mathfrak{h}$ invariant under the adjoint action of $G$ on $\mathfrak{g}$?
I know that this is true if $H$ is a normal subgroup. It is also true if $p = 0$, because then there is a bijection between connected normal subgroups and ideals of the Lie algebra.
What about when $p > 0$?
No. Let $k$ be a field of characteristic $p$ and define a group structure on $\mathbb{A}^2_k$ by $$(x_1, x_2) \ast (y_1, y_2) = (x_1+y_1, x_2+y_2 + x_1 y_1^p)$$ $$(x_1, x_2)^{-1} = (-x_1, -x_2 + x_1^{p+1})$$ We have $$[(x_1, x_2), (y_1, y_2)] = (x_1,x_2) (y_1,y_2) (-x_1, -x_2 + x_1^{p+1}) (-y_1,-y_2+y_1^{p+1})$$ $$=(0,x_1^p y_1 -x_1^p y_1 ).$$ So the commutator is trivial to second order, and the Lie algebra is abelian. Therefore, every subspace is an ideal.
We compute the adjoint action: $$(x_1,x_2) (y_1, y_2) (-x_1, -x_2+x_1^{p+1}) = (y_1, y_2 + x_1 y_1^p - y_1 x_1^p) \equiv (y_1,y_2-x_1^p y_1) \bmod \langle y_1,y_2 \rangle^2.$$ So $$Ad((x_1,x_2)) = \begin{bmatrix} 1 & 0 \\ -x_1^p & 1 \end{bmatrix}.$$ and we compute that (for example) the ideal $(t,0)$ is not fixed by the adjoint action.