Let $G$ be a semisimple Lie group, i.e. $G$ is connected and Lie algebra of $G$ is semisimple. We know by Iwasawa decomposition, there are connected subgroups $K$, $A$ and $N$ of $G$ such that the multiplication map from $K\times A\times N$ to $G$ is a smooth diffeomorphism. It is well known that $A$ and $N$ are simply connected and Lie algebra of $A$ is abelian. Since Lie algebra of $A$ is abelian and $A$ is connected and simply connected, clearly $A$ is isomorphic to $\mathbb R^n.$ This means that $G$ contains a copy of $\mathbb R^n.$ Moreover, $\{1\}\times A\times\{1\}=A$ is a closed subgroup of $G.$ This shows that any semisimple Lie group is noncompact if $A\neq \{1\}$ !! I am new in this business. Is my claim right? What can be necessary and sufficient condition for compactness of a semisimple Lie group in this direction of thoughts?



Iwasawa decompostion and simply connected subgroups
As you say, $K$ is compact just when $G$ has finite center. We can see this in the case of the universal covering group of the isometry group of the hyperbolic plane. Connected and simply connected solvables are diffeomorphic to Euclidean space. So for any semisimple Lie group $G=KAN$, we see that $A=1$ and $N=1$ and $K$ has finite center just when $G$ is compact, in which case $G=K$.


