If one has a normal Lie group inclusion $H\to G$, with quotient $G/H$, and a $G$-manifold $X$, one can take the quotient $X/H$. Then the $G$-action on $X/H$ factors thru a $G/H$-action, so one can take the quotient again to obtain $(X/H)/(G/H)$. It is known classically that $(X/H)/(G/H)\cong X/G$ (see, e.g. Bourbaki's book on Lie Groups and Lie Algebras, Section 1.6 of Chapter III). 


In a less structured setting, say, May's "Classifying Spaces and Fibrations" where the $H$ and $G$ are not Lie groups, but only topological monoids, and $X$ is just an object of the category of compactly generated Hausdorff spaces, we can still produce the quotient monoid $G/H$ and quotient space $X/H$ by a bar construction.


One could take this one step further and work just with a spectrum $X$ with an action of a topological monoid $G$. One can still take the quotient by $H$, and can still try to then take the quotient again by $G/H$. This all seems like it should be some purely formal bar construction thing.


My question is: where (if anywhere) can I find a proof that for a general model or quasi- category $C$, and a $G$-action on an object $X\in C$, the iterated quotient $(X/H)/(G/H)$ is equivalent to $X/G$? 


It'd also be nice to see a general proof for spaces or spectra.



Iterated Homotopy Quotient
Here's a sketch. The sequence of maps to look at is actually


$$BH \to BG \to B(G/H).$$


Recall that an object $X$ with a $G$-action in an $\infty$-category $C$ is the same thing as a functor $BG \to C$. I claim that the left Kan extension of this to a functor $B(G/H) \to C$ has the effect of quotienting by $H$ and then remembering the $G/H$-action; this follows from the fact that the above is a fiber sequence, and that left Kan extensions take "fiberwise colimits." Then the further quotient by the $G/H$ action is just taking the further left Kan extension along the unique map $B(G/H) \to \bullet$. 


On the other hand, taking the left Kan extension along $BG \to \bullet$ has the effect of quotienting by $G$. So the desired result follows from the observation that Kan extensions compose. 


