2 clarification

"1.That the isometry group of a Riemannian manifold is always a lie group."
This is the famous Myers-Steenrod theorem, proven in 1939 (Myers, S.B. and N.E. Steenrod: The group of isometries of a Riemannian manifold. The Annals of Mathematics, Vol 40, No. 2, April 1939, p. 400-416.)
It is in fact highly non-trivial, and I think you need that the manifold is connected

Your point "3.That isometry group of a Riemannian manifold is compact IFF the Riemannian manifold is compact." is as Igor pointed out false, the only thing which is right is the following

3.If the (connected) Riemannian manifold is compact then the isometry group is compact.
This is also a part of Myers-Steenrod theorem, and can be found in the reference above.

The "idea" of the proof is the following: (Let $(M,g)$ be a Riemannian manifold)

• Show that $(G=Iso(M,g), CO, op)$ is a locally compact topological transformationgroup.Here $CO$ is the compact-open topology, and $op: G \times M \rightarrow M$ the group action. Moreover $(M,g)$ compact implies $(G, CO, op)$ compact.
• Show that any tangential subgroup $H$ of $Diff(M)$ inherits a differentiable structure $[b]$ such that $(H,[b],op)$ ($op$ being the natural operation on $M$) is a Lie-Transformation group which is first-countable. The underlying topology $\tau$ is finer than $CO$-topology.
(If $(M,g)$ has countable many connected components, $G$ is a tangential subgroup of $Diff(M)$)
• Show that the topology $\tau$ cannot be strictly finer than the $CO$-topology. (needs frame-bundles, etc.)
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