Higman's criterion characterizes relative projectivity through the relative trace idempotent test

Let $\epsilon :{\mathrm{Ind}}_H^G{\mathrm{Res}}_H^{G}M\to M$ be the adjunction counit, written on the transversal model as $\epsilon (f)={\sum }_{t\in T}t\cdot f(t)$. By [L1] and [L2], a $kG$-module is relatively $H$-projective exactly when this counit splits.

Suppose first that $M$ is relatively $H$-projective. By step 1.1 choose a $kG$-map $s:M\to {\mathrm{Ind}}_H^G{\mathrm{Res}}_H^{G}M$ with $\epsilon s={\mathrm{id}}_M$. Define $\alpha (m):=s(m)(1)$. For $h\in H$, one has $s(hm)(1)=h\cdot s(m)(1)$ because $s$ is $G$-equivariant, so $\alpha \in {\mathrm{End}}_{kH}(M)$. Then ${\mathrm{Tr}}_H^G(\alpha )(m)={\sum }_{t\in T}t\cdot s(t^{-1}m)(1)={\sum }_{t\in T}t\cdot s(m)(t)=\epsilon (s(m))=m$. Hence ${\mathrm{id}}_M={\mathrm{Tr}}_H^G(\alpha )$.

Conversely, suppose ${\mathrm{id}}_M={\mathrm{Tr}}_H^G(\alpha )$ for some $\alpha \in {\mathrm{End}}_{kH}(M)$. Define $s(m)(g):=\alpha (g^{-1}m)$ for $g\in G$. Because $\alpha$ is $H$-linear, $s(m)(gh)=h^{-1}s(m)(g)$, so $s(m)$ lies in the induced module. The rule $m\mapsto s(m)$ is $G$-equivariant. Applying the counit of step 1.1 gives $\epsilon (s(m))={\sum }_{t\in T}t\cdot \alpha (t^{-1}m)={\mathrm{Tr}}_H^G(\alpha )(m)=m$. So $s$ splits the counit, and step 1.1 makes $M$ relatively $H$-projective.

Steps 2.1 and 3.1 prove the equivalence. For $H=G$, the transversal is $\{1\}$ and the trace condition reduces to $\alpha ={\mathrm{id}}_M$, as expected.