# Identity component of the normalizer is the centralizer

Let $G$ be a connected algebraic group and $H$ a closed connected subgroup of $G$ (these are the general hypothesis I work with).

Let us denote $N_G(H) = \{ g \in G \, / \, \forall h \in H; \, ghg^{-1} \in H \}$ the Normalizer of $H$ in $G$ and $C_G(H) = \{ g \in G \, / \, \forall h \in H; \, gh = hg \}$ the Centralizer of $H$ in $G$.

Is it true that $C_G(H)$ is the identity component of $N_G(H)$ ? (i.e. the greatest connected algbraic group that contains the identity element).

• No it is not in general. Take $G=H=SL_2$ $N_G(H)=G$ is connected but $C_G(H)=Z(G)$ is of order $2$ and thus disconnected. You might want to look for more hypothesis in your problem. – Clément Guérin Apr 5 '18 at 23:30
• What if, for example, $H$ is a proper subgroup of $G$? – Kal_Aki Apr 5 '18 at 23:44
• For example, I think it is possible if $G = GL_n(\mathbb{k})$ and $H = D_n(\mathbb{k})$. Is it right? – Kal_Aki Apr 6 '18 at 0:29
• In the case you mention it works. More generally if you take $H$ to be a Cartan subgroup of a connected reductive group $G$ it will work as well. – Clément Guérin Apr 6 '18 at 1:10