Relative Frobenius Structure on the Category of G-modules Let $G$ be a group $H\leq G$ a subgroup of finite index. Further, let ${\mathcal E}^G_H$ denote the class of those short exact sequences of $G$-modules (over some fixed base ring) which split when regarded as sequences of $H$-modules. Then $(G\text{-mod},{\mathcal E}^G_H)$ is a Frobenius category.
Does anybody know if it is true that $(G\text{-mod},{\mathcal E}^G_H)$ is not Frobenius if $(G:H)=\infty$?
To show this, it would be enough to construct for and pair $(G,H)$ with $(G:H)=\infty$ an $H$-module $M$ such that the canonical $H$-split map 
$\text{Ind}^G_H M\hookrightarrow\text{Coind}^G_H\text{Res}^G_H\text{Ind}^G_H M$
is not $G$-split. For example, if $M$ is the trivial $H$-module, this map becomes
${\mathbb Z}[G/H]\hookrightarrow\text{Hom}_{{\mathbb Z}H}({\mathbb Z}G,{\mathbb Z}[G/H])$
$g_0 H\longmapsto (g\mapsto g g_0 H)$.
Unfortunately, I'm not able to proof that this map is not $G$-split.
 A: If you take $G$ infinite and $H=(1)$ trivial, you are asking whether the category $G$-mod, with the class of short exact sequences which split in the category of abelian groups, is Frobenius. In that category $\mathbb ZG$ is projective, yet it is not injective---indeed, the relative $\mathrm{Ext}$-groups $\mathrm{Ext}_{\mathbb ZG,\mathbb Z}^p(\mathbb Z,\mathbb ZG)$ are just the usual $\mathrm{Ext}$'s, and they are not zero in general.
A: The representation $\mathrm{Hom}_{H}(\mathbb{Z}G,\mathbb{Z}[G/H])$ is the $\mathbb Z$-valued functions on the $G$-set $G\times_HG/H$ such that the each fiber of the natural map $G\times_HG/H\to G/H$ is only has finitely many non-zero entries.  
Now, what would a map of this representation to the trivial one look like?  It would have to send the characteristic function of  at least one element $\delta_x$ to a non-zero value.  But, take a transversal $t:G/H\to G$ and consider $\sum_{k\in G/H}t(k)\cdot \delta_x$.  This is a perfectly good element of $\mathrm{Hom}_{H}(\mathbb{Z}G,\mathbb{Z}[G/H])$, but it is sent to $\infty$ by our coinvariant. Thus, the coinvariants are trivial.  The coinvariants of $\mathbb{Z}[G/H]$ are not trivial, and any inclusion which is not injective on coinvariants does not split.
So indeed this exact structure is Frobenius if and only if the index is finite.
