action of $H\times K$ on $G$ given by $(h,k).g=hgk^{-1}$ depends only on ${\mathfrak g}$, ${\mathfrak h}$, ${\mathfrak k}$, and the inclusions ${\mathfrak h}\hookrightarrow {\mathfrak g}$ and $ {\mathfrak k}\hookrightarrow  {\mathfrak g}$.

Finally, we mention a result that shows how to deal with the situation when $H$ is not connected.

\[redcon\] Assume $K$ is connected and let $H_0$ be the identity component of $H$. If the $K$-action on $G/H_0$ is Cohen-Macaulay, then so is the $K$-action on $G/H$.

Use the characterization given by Prop. \[crit\], (c). Note that $H_0\times K$ is the identity component of $H\times K$, hence $H^*_{H\times K}(G)\simeq H^*_{H_0\times K}(G)^{H/H_0}.$ By hypothesis, $H^*_{H_0\times K}(G)$ is Cohen-Macaulay as a module over the ring $H^*(BH_0 \times BK)$, thus also over its subring $H^*(BH\times BK)=H^*(BH_0 \times BK)^{H/H_0}$, since the ring is a finitely generated module over the subring and one can use Lemma \[serre\]. By a standard averaging argument, the invariant module $H^*_{H_0\times K}(G)^{H/H_0}$ can be realized as a direct summand of $H^*_{H_0\times K}(G)$ in the category of $H^*(BH \times BK)$-modules, hence it is Cohen-Macaulay.

Hermann actions {#semisimple}
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In this section we will prove Thm. \[main\]. The meaning of $G$, $H$, and $K$ is as in Sect. \[intro\]. In particular, the 