Maximal compact subgroups of $GL_n(\mathbb{R})$. The subgroup $O_n=\{M\in GL_n(\mathbb{R}) | ^tM M = I_n\}$ is closed in $GL_n(\mathbb{R})$ because it's the inverse image of the closed set $\{I_n\}$ by the continuous map $X\mapsto ^tX X$. $O_n$ is also bounded in $GL_n(\mathbb{R})$, for example this is clear by considering the norm $||X|| = \sqrt{tr(^tX X)}$ (elements of $O_n$ are bounded by $\sqrt{n}$), so $O_n$ should be a compact subgroup of $GL_n(\mathbb{R})$. 
I see it claimed without proof in many places that $O_n$ is a maximal compact subgroup. How can we see this?

Suppose $G\supseteq O_n$ is a compact subgroup of $GL_n(\mathbb{R})$. Let $A\in G$ and $A=U\Sigma V^T$ be a singular value decomposition of $A$. Since $O_n\subseteq G$, it follows that $\Sigma\in G$ and in turn $\Sigma^k\in G$ for $k=1,2,\ldots$. If $\sigma_1(A)>1$, then $\{\Sigma^k\}_{k=1}^\infty$ and hence $G$ is unbounded, which contradicts the boundedness of $G$; if $1=\sigma_1(A)=\cdots=\sigma_r(A)>\sigma_{r+1}(A)$ for some $r<n$, then $\lim_{k\to\infty}\Sigma^k=I_r\oplus0\notin G$, which contradicts the closedness of $G$. Therefore $\Sigma=I$ and $A=UV^T\in O_n$, i.e. $G\subseteq O_n$.
