In Springer, Invariant Iheory, page 7, it is claimed that for a group $G \subset GL(V)$, and $\overline G$ its closure, we have $S^G = S^{\overline G}$. (Here $S = k[t_1, \dots, t_n]$ where $n= \dim V$).
He said that it follows from the fact that $\overline G$ is a group if $G \subset GL(V)$ is, but I don't see how to use it. I have no ideas about it. Any hints ?
Let $H$ be a closed subgroup of $\textrm{GL}_n$, $X$ an affine variety, and $H \times X \rightarrow X$ a group action which is a morphism of varieties. Then for $h \in H$, you obtain a $k$-algebra isomorphism $\lambda_x: k[X] \rightarrow k[X]$ given by $\lambda_h(f)(x) = f(h^{-1}.x)$, where $k[X]$ is the coordinate ring of $X$.
The assignment $x \mapsto \lambda_x$ defines a group homomorphism $H \rightarrow \textrm{GL}(k[X])$, where $\textrm{GL}(k[X])$ is the group of vector space isomorphisms of $k[X]$.
In your case, $X = k^n$, and $k[X] = S = k[T_1, ... , T_n]$, and $H$ acts on $X$ by matrix multiplication.
Lemma: Let $s \in k[X]$. There exists a finite dimensional subspace $W$ of $k[X]$ containing $s$ which is $H$-stable, in the sense that $\lambda_x(f) \in W$ for any $f \in W$. Moreover, choosing a basis for $W$ and identifying $W$ with affine $m$-space for some $m \geq 1$, the function
$$H \times W \rightarrow W$$ $$(h,f) \mapsto \lambda_h(f)$$ is a morphism of varieties.
Proof: Any book on linear algebraic groups. $\blacksquare$
Now take $H = \textrm{GL}_n, X = k^n$, $s \in S^G$. Find a finite dimensional subspace $W$ containing $s$ as in the lemma which is stable under all $\lambda_h : h \in \textrm{GL}_n$. The function $$\phi: \textrm{GL}_n \rightarrow W$$ $$\phi(h) = h.s = \lambda_h(s)$$ is a morphism of varieties and hence continuous, as it is obtained from a morphism of varieties $\textrm{GL}_n \times W \rightarrow W$. The image of $G$ consists of the singleton set $\{s\}$, which is closed in $W$ (one point sets are closed in affine space). But since $\phi$ is continuous,
$$\phi(\overline{G}) \subseteq \overline{\phi(G)} = \overline{ \{s\}} = \{s\}$$ so $s$ is also fixed by $\overline{G}$. Thus $S^G \subseteq S^{\overline{G}}$.