Algebraic subgroups $ H $ of $ G $ such that $ H(k) = G(k) $

Question

The following theorem can be found in Milne's book on algebraic groups (Theorem 1.45) and in his online course notes (Proposition 1.31):

Theorem: Let $G$ be an algebraic group (= algebraic group scheme, not necessarily affine) over a field $ k $ and $S$ a closed subgroup of $G(k)$. Then there is a unique reduced algebraic subgroup $H$ of $G$ such that $ H(k) = S$.

From this I deduce:

Corollary: Let $G$ be a reduced algebraic group over a field $ k $ and let $H$ be a reduced algebraic subgroup of $G$ such that $ H(k) = G(k) $. Then $ H = G $.

Proof: Apply the Theorem with $ S = G(k) $.

However:

Counterexample to the Corollary: Let $k$ be the field of real numbers and let $G=\mu_3$ be the group of 3rd roots of unity (i.e. the Hopf algebra of $G$ is $k[X]/(X^3-1)$). Since $k$ is of characteristic zero, $G$ is smooth and thus reduced. Let $ H $ be the trivial subgroup of $G$ (i.e. $H(R) = \{1\} $ for all $k$-algebras $R$). Then $ H(k) = \{1\} = G(k) $ and $G,H$ are both reduced, but $G \neq H$.

Where is my mistake?

(If it helps, I only care about affine group schemes.)

Answer 1

The deduction in the question and the example are correct, but the Theorem is wrong. This is listed in the Errata for the book. The correct version of the theorem should be that, under the same assumptions, there exists a unique reduced algebraic subgroup $H$ of $G$ such that $ |H| = \overline{S} $, where $|H|$ is the underlying topological space of $H$ and $ \overline{S} $ is the Zariski closure of $S$ in $G$. This algebraic subgroup $H$ satisfies $ H(k) = |H| \cap G(k) = \overline{S} \cap G(k) = S $, but it is not necessarily uniquely determined by the property $ H(k) = S $. Thus the Corollary stated in the question is not correct in this general form.