Commutativity of induction and inflation of representations Let $G$ be a finite group, $H$ a subgroup and $N$ a normal subgroup. Let $\chi$ be a representation of $HN/N$. Does $$Ind_{HN}^G Inf_{HN/N}^{HN} \chi = Inf_{G/N}^G Ind_{HN/N}^{G/N} \chi$$ always hold? Here $Ind$ and $Inf$ are the induction and inflation functors. 
If this isn't true, are there any simple sufficient conditions for it to hold (possibly with a sketch proof/reference)?
 A: Let $\tau$ be an irreducible subrepresentation of $Ind_{HN}^GInf_{HN/N}^{HN}\chi$, so by Frobenius reciprocity there's a non-zero map $\tau|_{HN}\rightarrow Inf_{HN/N}^{HN}\chi$. So $N$ acts trivially through $\tau$, meaning that we can identify $\tau|_{HN}$ with a representation of $HN/N$, and doing so we get a non-zero map $\tau|_{HN}\rightarrow\chi$. Using Frobenius reciprocity again, $\tau$ is a subrepresentation of $Ind_{HN/N}^{G/N}\chi$, and inflating gives a non-zero map $\tau\rightarrow Inf_{G/N}^GInd_{HN/N}^{G/N}\chi$. So every irreducible subrepresentation of the left hand side is a subrepresentation of the right hand side. That means that the only way that the two representations can't be equal is if some subrepresentations appear with different multiplicities, but since $|G/HN|=|G|/|HN|=|G||H\cap N|/|HN|=|G/N|/|H/H\cap N|=|G/N|/|HN/N|$, you can compare dimensions to conclude that the representations genuinely are equal.
A: Yes, it holds (as long as your equality sign stands for canonical isomorphism). See Exercise 4.1.11 in Darij Grinberg and Victor Reiner, Hopf algebras and combinatorics, 11 May 2018, arXiv:1409.8356v5 (see the ancillary file for the solution). Note that our $K$, $H$ and $G$ correspond to your $N$, $HN$ and $G$.
