The closure of a subgroup of GL_n is a subgroup

Idea. The group operations are continuous, so they pass to limits.

Closure by sequences. $GL_n(\mathbb{R})$ is a metric space (subset of $\mathbb{R}^{n^2}$), so $\overline H$ is the set of limits in $GL_n$ of sequences from $H$.

Subgroup test. $\overline H$ is nonempty ($I\in H$). Let $a,b\in\overline H$ and choose $a_k,b_k\in H$ with $a_k\to a$, $b_k\to b$ in $GL_n$. Then:
- $b_k^{-1}\to b^{-1}$ (inversion is continuous on $GL_n$: entries of $B^{-1}$ are rational functions of those of $B$, with denominator $\det B\ne0$);
- $a_kb_k^{-1}\to ab^{-1}$ (multiplication is continuous).

Since $a_kb_k^{-1}\in H$, the limit $ab^{-1}$ lies in $\overline H$. So $\overline H$ is closed under $(a,b)\mapsto ab^{-1}$, and $\overline H$ is a subgroup.

Warning. The closure must be taken in $GL_n$. The closure in $M_n(\mathbb{R})$ may contain singular matrices: e.g. $H=\{\lambda I:\lambda>0\}$ has closure $\{\lambda I:\lambda\ge0\}$ in $M_n$, which contains $0$ and is not a group.

Remark. If $H$ is normal, so is $\overline H$ (conjugation is continuous), and if $H$ is abelian then $\overline H$ is abelian.