Let $G$ be a semi-simple real Lie group such that $|\pi_0(G)|<\infty$ and let $K$ be a maximal compact subgroup of $G$.

Q1: How does one prove that $N_G(K)=K$?

So I know a nice (and low-tech) proof of this result in the special case where $G=SL_n(\mathbf{R})$. Let $K=SO_n(\mathbf{R})$ (a maximal compact). Then the associated symmetric space of $G$, namely, $D=G/K$ can be thought of as the set of positive definite symmetric matrices of determinant $1$ (which is the same thing as the set of positive definite quadratic forms of determinant $1$). Now let $Q_K(x)\in D$ be the standard quadratic form with isotropy group equal to $O_n(\mathbf{R})$ (inside the full group $GL_n(\mathbf{R})$). Then if $g\in N_G(K)$, we see directly that the isotropy group of $Q_K(g^{-1}x)$ has to contain $SO_n(\mathbf{R})$ and therefore has to be equal to $O_n(\mathbf{R})$. However, if two non degenerate quadratic forms in characteristic zero have the same isotropy group then they they differ by a non-zero scalar (see this link for a proof). Since the determinant of $g$ is equal to $1$ then the scalar has to be $1$. Thus $Q_K(x)=Q_K(g^{-1}x)$ and therefore $g\in K$.

Q2: Is it possible to generalize the proof above in an obvious way to an arbitrary semi-simple Lie group?

I don't quite see how to use the semi-simplicity of $G$. Note that I would prefer to avoid , if possible, the existence of the Iwasawa decomposition.

reductive(!), so $N_G(K^0)/K^0 = L(\mathbf{R})$ for a reductive $\mathbf{R}$-group $L$ and $L^0(\mathbf{R})$ has no non-trivial connected compact subgroups. Thus, $L^0$ has no relative roots ($S^1\subset{\rm{SL}}_2$),so $L^0$ is anisotropic and hence $L^0(\mathbf{R})$ is compact, so $L^0=1$. QED $\endgroup$ – user30180 Jan 18 '13 at 1:53