TITLE: homotopy type of connected Lie groups
QUESTION [20 upvotes]: Is there a simple proof (short and low-tech) of the following fact:
(E. Cartan) A connected real Lie group $G$ is diffeomorphic (as a manifold) to 
$K\times\mathbb{R}^n$ where $K$ is a maximal compact subgroup of $G$. Moreover, 
all maximal compact subgroups of $G$ are conjugate to $K$.
One nice corollary of this is that the group $K$ is a deformation retract of $G$ so these
two groups have the same homotopy type. In the case where $G$ is semi-simple one may deduce the first part of the result above using the existence of the Iwasawa decomposition which is not so trivial to prove! But here since I'm only asking for a diffeomorphism between two manifolds, there might be a simpler argument!

REPLY [2 votes]: Yet another traditional favorite argument, in reality a variant on the more geometric argument of Cartan's, that applies immediately to $G=GL(n,\mathbb R)$ and $K=O(n,\mathbb R)$ uses the model $G/K\approx C$, the cone of positive-definite symmetric $n$-by-$n$ matrices, via a Cartan decomposition $G=C\cdot K$, which itself follows from the spectral theorem for symmetric operators.
Then, given a compact $K'\subset G$, pick any $x_o\in C$, and consider the average $x=\int_{K'} k\cdot x_o\;dk$. Since the cone $C$ is non-degenerate, $x\not=0$, and, by design, it is $K'$-invariant. Since $G$ is transitive on $C$, all isotropy groups of points are conjugate to the isotropy group of $1_n$, namely, $O(n,\mathbb R)$.
This idea can be adapted to some other of the classical groups.