Maximal compact subgroup

Remarks

Example: $K = \mathrm{O}(n)$ in $G = \mathrm{GL}_n(\R)$. It is closed and bounded in the space of matrices, so it is compact. Let $L \supseteq \mathrm{O}(n)$ be a compact subgroup. Averaging the standard inner product over $L$ with the Haar measure gives an inner product $\ip{\cdot}{\cdot}'$ that $L$ preserves. The group $\mathrm{O}(n)$ preserves it too, and the only $\mathrm{O}(n)$-invariant inner products are the positive multiples of the standard one. So $L$ preserves the standard inner product, and $L \subseteq \mathrm{O}(n)$.

Non-example: $\mathrm{SO}(n)$ in $\mathrm{GL}_n(\R)$ is compact, but it fails clause 2, since $\mathrm{SO}(n) \subsetneq \mathrm{O}(n)$.

A maximal compact subgroup need not exist. In the countable direct sum $\bigoplus_{i \in \N} \Z/2$ with the discrete topology, the compact subgroups are the finite ones, and each finite subgroup lies in a larger finite subgroup.