In this delightful question, the poster mentioned that the isometry group of a compact Lie group $G$, equipped with the metric from the Killing form, is $G\times G/Z(G)$, where $Z(G)$ is the center of $G$. I'm specifically interested in the Riemannian case, so I will add the extra condition that $G$ be semisimple.

Is there a concise way to show that $G\times G/Z(G)$ is the isometry group, other than using the algorithm from the answer?

I can clearly see that it is a subgroup, since the metric coming from the Killing form is bi-invariant, and that quotienting out the center from one of the two copies of $G$ is necessary, because otherwise the action wouldn't be faithful. However, I cannot see how all isometries are of that form. In fact, I'm almost positive that this would exclude something of finite order, like $g\mapsto g^{-1}$.

identity componentof the isometry group. It is not surprising that finite-order isometries live in different components! $\endgroup$ – LSpice Dec 29 '14 at 1:05the Lie algebraof the group of isometries of a homogeneous space $G/H$ endowed with a $G$-invariant (pseudo-)Riemannian metric $g$" (emphasis mine). That is, he does explicitly mention (though I missed it at first!) that he is only identifying the Lie algebra; and the Lie algebra of a Lie group, which explicitly 'zooms in' near the identity, only ever has a chance of identifying the identity component. $\endgroup$ – LSpice Dec 29 '14 at 1:31