In Milne's book "Algebraic Groups: the theory of group schemes of finite type over a field. Cambridge", Chapter 8, proposition 8.9:
X is also reduced
In second paragraph of the proof, why we can get $$\mathcal{I}\mathcal{O}_p \subset \mathfrak{m}^n_p$$
In the Corollary 8.10 which without assumption of $G$ is reduced:
Apply the proof to the faithful action $G/Z \times G \to G$ by conjugation, where $Z$ is centre, we get that $$G_{red} \subset G^H \subset G$$ where $H$ algebraic subgroups of $G/Z$.
We can also identity $G^H = C_G(H^{\prime})$, where $H^{\prime}$ algebraic subgroup of $G$ correspondence to $H$.
How can I get $H = {e}$ or $G^H = G$ or $H^{\prime} \subset Z$?
Appreciated for any help!
Assume $X$ is reduced in 8.9. Replace the proof of 8.10 with: Certainly, the kernels contain $Z(G)$. As in the proof of 8.9, there exists an algebraic subgroup $H$ of $G$ such that $H$ is the kernel of $\rho_{n}\colon G\rightarrow\textrm{GL}_{\mathcal{O}_{e}/\mathfrak{m}_{e}^{n+1}}$ for all sufficiently large $n$. Moreover, $G^H$ (which equals $C_G(H)$) contains an open neighbourhood $U$ of $e$. Now $C_G(H)$ contains $U\cdot U$, which equals $G$ (Exercise 1-2), and so $C_G(H)=G$. Therefore, $H\subset Z(G)$.