Let $G$ be a linear algebraic group and $C_G(x)$ be the centralizer of $x$. Show that $C_G(x)$ is a closed subgroup for all $x\in G$.

Question

Let $G$ be a linear algebraic group and $C_G(x)$ be the centralizer of $x$. Show that $C_G(x)$ is a closed subgroup for all $x\in G$.

I want to enstablish a homomorphism from $G$ to $G$, and the kernel is $C_G(x)$, but I can't find it.

Answer 1

Let $g_nx = x g_n$ for all $g_n \in G$. If $g_n \rightarrow g$ then $g_nx \rightarrow gx$ and $x g_n \rightarrow xg$ and since $g_n x = x g_n$, we have that $xg = gx$. Hence $g \in C_G$. Hence $C_G(x)$ is a closed set.