Let $k$ be an algebraically closed field and let $G = \operatorname{GL}_{n,k}$. It can be easily proven that the normalizer of a closed subgroup of $G$ is again a closed subgroup of $G$. But now suppose $H\leq G$ is an abstract subgroup of $G$, i.e., it is not necessarily closed. Then is it still the case that the normalizer $N_G(H)$ is a closed subgroup?
As a counterexample, I thought of $H=D_{n,\mathbb Q}$ diagonal matrices with non-zero rational entries. Then the normalizer contains generalized permutation matrices with rational entries and the center of $\operatorname{GL}_{n,k}$, $\{\lambda I_n : \lambda\in k^*\}$. I am not sure if this is all but is this already enough to argue that the normalizer is not closed?
I came up with a counter-example and forgot to post it here, so here goes: For completeness, let us first show that this holds if $H$ is a closed subgroup of $G$. Let $x\in G$ and define $\phi_x : G\to G$, $g\mapsto gx$. We have that: \begin{align} N_G(H) &= \{g\in G : gHg^{-1} = H \}\\ &=\{g\in G : \forall h\in H\hspace{10px}\phi_h(g) \in H \}\\ &= \bigcap_{h\in H} \phi_h^{-1}(H) \end{align} As $\phi_x$ is a morphism of varieties, it is continous and hence $H\subseteq G$ being closed, we have that $\phi_h^{-1}(H)\subseteq G$ is closed for any $h\in H$. Since arbitrary intersection of closed subsets are closed, the result follows.
Now, as per the counter-example, let $k=\mathbb C$, $G=\operatorname{GL}_{n,\mathbb C}$ and $H = B_{n, \mathbb Q(i)}$ the subgroup of upper triangular matrices with entries in $\mathbb Q(i)^\times$ where $\mathbb Q(i) = \{a+bi : a,b\in\mathbb Q\}$ is the field of Gaussian rationals. Then $N_G(H)$ is contained in $N_G(B_{n,\mathbb C})$ as a dense subset. This is because it is contained as a dense subset in the standard topology and Zariski topology being coarser than the standard topology, it must be a dense subset in the Zariski topology as well. We further have that $B_{n,\mathbb C}$ is self-normalizing and since it is closed, it is enough to show that $N_{\operatorname{GL}_{n,\mathbb C}}(B_{n, \mathbb Q(i)})\neq B_{n, \mathbb C}$. To illustrate this, consider the following simple example in the case of $n=2$: $$ X = \begin{pmatrix} \sqrt{2} & 1 \\ 0 & 1 \end{pmatrix} \in B_{2,\mathbb C}, \qquad Y = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \in B_{2,\mathbb Q(i)} $$ \begin{align} \implies XYX^{-1} &= \begin{pmatrix} \sqrt{2} & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ 0 & 1 \end{pmatrix} \\ &= \begin{pmatrix} \sqrt{2} & \sqrt{2} + 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ 0 & 1 \end{pmatrix} \\ &= \begin{pmatrix} 1 & \sqrt{2} \\ 0 & 1 \end{pmatrix} \notin B_{2, \mathbb Q(i)} \end{align} Therefore $X\notin N_{\operatorname{GL}_{2,\mathbb C}}(B_{2, \mathbb Q(i)})$ and $N_{\operatorname{GL}_{2,\mathbb C}}(B_{2, \mathbb Q(i)})\neq B_{2, \mathbb C}$. This can be generalized to higher $h$ by appending columns of $1$s to the right of both $X$ and $Y$.
Hence, $N_{\operatorname{GL}_{n,\mathbb C}}(B_{n, \mathbb Q(i)})\neq B_{n, \mathbb C} = N_{\operatorname{GL}_{n,\mathbb C}}(B_{n, \mathbb C}) = \overline{ N_{\operatorname{GL}_{n,\mathbb C}}(B_{n, \mathbb Q(i)}) }$ and the normalizer of $B_{n,\mathbb Q(i)}$ in $\operatorname{GL}_{n,\mathbb C}$ is not closed albeit a subgroup.