I am studying Differential Geometry and I am facing with a lemma in which there is a step that I do not understand. In particular, let $G$ be a connected compact Lie group, is used "$\langle\ \cdot , \cdot \rangle$ any $Ad(G)$-invariant scalar product (or equivalently a non-degenerate semi-positive bilinear form) on $\mathfrak{g}$, the Lie algebra of $G$ ". I can't find any such of scalar product.
When a connected compact Lie group $G$ has such a scalar product? I tried with the Killing form that is semi-definite negative bilinear form, that is non-degenerate iff the lie algebra is semisimple, but how I can prove that a compact connected Lie group have semisimple Lie algebra?
I tried also with a proposition (3.6.2) in "lie groups", Duistermaat-Kolk, saying that the killing form of a lie algebra $\mathfrak{g}$ is negative on the complement of the center $\mathfrak{z}$ of $\mathfrak{g}$. Then if I prove that $G$ is centerless, so is $\mathfrak{g}$ and the opposite of the killing form is an $Ad(G)$-invariant scalar product on $\mathfrak{g}$.
$\underline{NOTE}$: The lemma I have to do states "Two maximal tori of a compact connected Lie group $G$ are coniugate", so we can suppose that $G$ is non-abelian.
A compact Lie algebra $\mathfrak{g}$ for a compact Lie group $G$ always admits an $Ad(G)$-invariant inner scalar product, see Propositioon $4.24$ here. A compact Lie algebra need not be semisimple in general, but it is always reductive, i.e., a direct sum of the semisimple commutator ideal and the center. Hence the Killing form is negative semidefinite - see also here for further references.