The outer automorphism group of a topological group $G$ is constructed by the short exact sequence
$$
1\longrightarrow \operatorname{Inn}(G) \longrightarrow \operatorname{Aut}(G) \longrightarrow \operatorname{Out}(G) \longrightarrow 1.
$$
This sequence does not always split, see [Non-split Aut(G) $\to$ Out(G)?](https://mathoverflow.net/questions/293436/), [for example](https://en.wikipedia.org/wiki/Outer_automorphism_group#In_symmetric_and_alternating_groups) for the discrete group $G = A\_6$.


I am interested in the case where $G$ is a compact, connected Lie group. Does the sequence always split in this case? (If $G$ has a simple Lie algebra $\mathfrak{g}$ then I believe the answer is [yes](https://en.wikipedia.org/wiki/Outer_automorphism_group#In_complex_and_real_simple_Lie_algebras).)



Does Aut(G) → Out(G) always split for a compact, connected Lie group G?
Yes, $\operatorname{Aut}(G) \to \operatorname{Out}(G)$ always splits. The proof is just as in [my answer](https://mathoverflow.net/a/378141) to your question [Classification of (not necessarily connected) compact Lie groups](https://mathoverflow.net/questions/377981/classification-of-not-necessarily-connected-compact-lie-groups): regard $\operatorname{Aut}(G)$ as an extension of $\operatorname{Inn}(G) = G/\operatorname Z(G)$ by a discrete group $\operatorname{Out}(G)$, and lift $\operatorname{Out}(G)$ to $\operatorname{Aut}(G)$ as the automorphisms that preserve a pinning in the sense of [that answer](https://mathoverflow.net/a/378141). (These are often called "diagram automorphisms".) Over in that other question we did not get an honest section of the component group inside the Lie group because you did not assume that the identity component was centreless, but since the adjoint group $\operatorname{Inn}(G)$ is centreless, everything is fine here.


