Quotient of induced modules Let $G$ be a finite group and $H \subset G$ a subgroup (if it simplifies the question, assume that $H$ is normal in $G$). Suppose that I have two $G$-modules, $M$ and $N$ such that  $M$ and $N$ are induced from $H,$ i.e., there are two $H$-modules $K$ and $L$ such that $M \cong \operatorname{Ind}_H^G K$ and $N \cong \operatorname{Ind}_H^G L.$ If I have a map of $G$-modules
$$f \colon M \rightarrow N,$$ is it true that $\operatorname{coker} f$ is also a $G$-module that is induced from $H?$
 A: The answer is no. Let $G$ be any nontrivial finite group, take $H=\{1\}$ to be trivial, and let $M\cong N\cong\mathbb{C}G$ be induced from the $1$-dimensional $H$-module. The map
$$
f:M\to N
$$ 
given by $f(v)=\sum_{g\in G}g.v$ is a $G$-module map, and the image of $f$ is the $1$-dimensional trivial submodule of $G$. Note that $N/\mathrm{im}(f)$ is not an induced module.
A: The example given by David Hill shows that it's false in all generality, however I'd like to point out that if $f$ itself is induced by a map $K\to L$, then the result becomes true. Indeed $\mathrm{Ind}_H^G$ is left adjoint to $\mathrm{Res}_H^G$ (restriction) and so it preserves colimits (as all left adjoints do). 
Hence if $f= \mathrm{Ind}_H^G (m)$, $m: K\to L$, then $\mathrm{coker}(m) : L\to L/\mathrm{im}(m)$ is the colimit of the obvious diagram, hence $\mathrm{Ind}_H^G(\mathrm{coker}(m)): N\to \mathrm{Ind}_H^G(L/\mathrm{im}(m))$ is the colimit of $f,0 : M\to N$, hence the cokernel of $f$ (up to a unique commuting isomorphism) . 
