Looking for example of quotient of group algebra by ideal of group ring which fails to be injective

Question

I am looking for an example of a group ring $\mathbb{Z}[G]$ of a finite group $G$ along with a lattice $I$ (in the case at hand the word 'lattice' means: a $\mathbb{Z}[G]$-submodule which is additionally a finitely generated free $\mathbb{Z}$-subalgebra [corrected]) such that \begin{equation} M:=\mathbb{Q}[G] / I \end{equation}

is $\textit{not}$ an injective $\mathbb{Z}[G]$-module.

Note that I really want $\mathbb{Q}[G]/I$ and not something like $\mathbb{Z}[G]/I$, where it is easy to produce non-injective examples (for example $I=0$ would do...).

To give some context: Since $\mathbb{Q}[G]$ is semisimple, every torsion-free divisible module is injective. This remains true for all divisible modules (now possibly having torsion) if and only if the order is hereditary - which group rings of finite groups practically never are.

So essentially every group ring of a finite group should have a torsion divisible module which fails to be injective. But do there exist such examples of the concrete shape of $M$ as above, i.e. as the quotient of the semisimple algebra itself?

Added in edit: If that helps: It would be equivalent to produce an example of a lattice $I$ inside $\mathbb{Z}[G]$ such that $\operatorname{Ext}_{\mathbb{Z}[G]}^2(N,I)\neq0$ for some $N$. (one sees this by looking at the $\operatorname{Ext}_{\mathbb{Z}[G]}^{\bullet}(N,-)$ long exact sequence of $I \hookrightarrow \mathbb{Q}[G] \twoheadrightarrow M$). For example, if I allowed infinite groups, then $G=\mathbb{Z}$ and $I$ the augmentation ideal would solve my question. However, I need a finite $G$...

Answer 1

So long as $G$ is nontrivial, the augmentation ideal of $\mathbb{Z}[G]$ still works.

If $I$ is any submodule of $\mathbb{Z}[G]$ then there is a short exact sequence of $\mathbb{Z}[G]$-modules $$0\to\mathbb{Z}[G]/I\to\mathbb{Q}[G]/I\to\mathbb{Q}[G]/\mathbb{Z}[G]\to0.$$ The last term is always injective, so if the middle term is injective, then $\mathbb{Z}[G]/I$ has injective dimension at most one, and so in particular $H^i(G,\mathbb{Z}[G]/I)=0$ for $i>1$.

But if $I$ is the augmentation ideal of $\mathbb{Z}[G]$ then $\mathbb{Z}[G]/I$ is just $\mathbb{Z}$ with trivial $G$-action, and there are arbitarily large $i$ with $H^i(G,\mathbb{Z})\neq0$.