\ldots, x_n)\\{}&=
(hx_2g_2^{-1}, g_2x_3g_3^{-1}, \ldots, g_{n-2}x_{n-1}g_{n-1}^{-1}, g_{n-1}x_n k^{-1}),\end{aligned}$$is Cohen-Macaulay. We continue the procedure and gradually drop out the $L$-factors in $H'\times L^{n-1}\times K'$ until we finally obtain the action of $H'\times K'$ on $L$ given by $(h, k)\cdot x=hxk^{-1}$. But this action is Cohen-Macaulay by the result already proved in Subsect. \[sec:simple\] (see also Prop. \[crit\]).

To deal with (ii), we start by modding out the action of the first factor $L$ in $H'\times L^{n-1}$, which is clearly a free action. In this way, we reduce the problem to showing that the action of $H'\times L^{n-2} $ on $L^{n-2}\times L/K'$ described by $$\begin{aligned}
{}&(h, g_2, \ldots, g_{n-1})\cdot(x_2, \ldots, x_{n-1}, x_nK')\\{}&=
(hx_2g_2^{-1}, g_2x_3g_3^{-1}, \ldots, g_{n-2}x_{n-1}g_{n-1}^{-1}, g_{n-1}x_n K'),\end{aligned}$$is Cohen-Macaulay. We continue the procedure until we are led to the action of $H'$ on $L/K'$ given by left translations. Again, this action is Cohen-Macaulay by the result we proved in Subsect. \[sec:simple\].

Similarly, in case (iii) we reduce the problem to the action of $L$ on $H'\backslash L \times L/K'$ described by $g\cdot (H'x_1, x_2K')=(H'x_1g^{-1}, gx_2K')$. The map $$H'\backslash L \times L/K'\to L/H'\times L/K', \quad (H'x_1, x_2K')\mapsto (x_1^{-1}H', K'x_2)$$ is an $L$-equivariant diffeomorphism. This allows us to change our focus to the action 