Higman's Criterion is a fundamental result in the representation theory of a finite group $G$ over a field $k$ of characteristic $p>0$. One version is the following (I hope the notation is standard enough):


Let $M$ be a $kG$-module and let $H$ be a subgroup of $G$. Then $M$ is a direct summand of ${\rm Ind}\_H^G\,{\rm Res}\_H^G\, M$ if and only if there is a $kH$-endomorphism $\psi$ of $M$ such that ${\rm tr}\_H^G(\psi)=1\_M$.


Question: suppose that $M$ has a $kH$-endomorphism $\psi$ such that ${\rm tr}\_H^G(\psi)=1\_M$. Does $\psi M$ contain a $kH$-submodule $L$ such that $L$ is a direct summand of ${\rm Res}\_H^G\,M$ and $M$ is a direct summand of ${\rm Ind}\_H^G\,L$?


Motivation: Notice that if $L$ is a direct summand of ${\rm Res}\_H^G\,M$ and if $N$ is a submodule of ${\rm Res}\_H^G\,M$ containing $L$, then $L$ is a direct summand of $N$. As a consequence, if $N$ is a submodule of ${\rm Res}\_H^G\,M$ which contains $\psi M$, then $M$ is a direct summand of ${\rm Ind}\_H^G\,N$. 
The standard argument proving the `if' implication in Higman's Criterion appears to show this consequence directly. This seems to me to be a mysterious property of $\psi M$, at least if my Question has a negative answer.


Going back to generalities, let $M$ be a $kG$-module and suppose that $M$ is a direct summand of ${\rm Ind}\_H^G\,N$, where $N$ is a $kH$-module. One version of Higman's Theorem produces a $kH$-endomorphism $\phi$ of $M$ such that $t\_H^G(\phi)=1\_M$. This $\phi$ `factors through' $N$ in the sense that it is a composition $\phi=\phi\_2\circ\phi\_1$ where $\phi\_1:{\rm Res}\_H^G\,M\rightarrow N$ and $\phi\_2:N\rightarrow{\rm Res}\_H^G\,M$. Set $L=\phi M$. Then $L$ is both a submodule and quotient module of ${\rm Res}\_H^G\,M$ and a subquotient ${\rm Im}(\phi\_1)/{\rm Im}(\phi\_1)\cap{\rm ker}(\phi\_2)$ of $N$ and $M\mid{\rm Ind}\_H^G\,L$.



Higman's Criterion
This has the rather easy answer: No!


Let $G$ be a finite group of order divisible by $p$ that has a $p$-block $B$ of defect $0$ and let $M$ be the unique simple $kG$-module in $B$. Take $H$ to be any nontrivial $p$-subgroup of $G$. Then $k\_H$ is a submodule of Res$\_H^G(M)$, as $k\_H$ is the unique simple $kH$-module. So $M$ is a submodule of Ind$\_H^G(k\_H)$, by Frobenius reciprocity. But $M$ is a projective $kG$-module. So $M$ is a direct summand of Ind$\_H^G(k\_H)$.


Now by the last paragraph in my posted question there is a $kH$-homomorphism $\phi$ of $M$ such that t$\_H^G(\phi)=1\_M$ and $\phi$ factors through $k\_H$. So $\phi M\cong k\_H$. But Res$\_H^G(M)$ is a projective $kH$-module and $k\_H$ is a non-projective $kH$-module (as $p$ divides $|H|$). So $\phi M$ has no submodule that is a direct summand of Res$\_H^G(M)$.


