I am in search of some examples of finite groups $G$ and modules over the ring $\mathbb{Z}[G]$ with following conditions (and examples taken independently). I do not have idea whether such examples are possible, and the question can be very trivial; but I didn't immediately get example(s) to justify some statements in an introduction to integral representations in Curtis-Reiner's representation theory book.

1) Finite group $G$ and a $\mathbb{Z}[G]$-module $M$ such that $M$ has no composition series. [This implies that Jordan-Holder theorem do not hold for $M$; am I right?]

2) Finite group $G$ and a $\mathbb{Z}[G]$-module $M$ such that Krull-Schmidt theorem fails for $M$.

3) Finite group $G$ and a $\mathbb{Z}[G]$-module $M$ such that $M$ contains a proper non-zero $\mathbb{Z}[G]$-submodule but has no complement (i.e. that submodule is not a direct summand).

If $\mathbb{Z}$ is replaced by a field $K$ whose characteristic does not divide $|G|$ then above examples are not possible- standard beginning of representation theory of groups. But Curtis-Reiner says that such results (Jordan-Holder/ Krull-Schmidt/ Maschke's theorem) are no longer true if the field $K$ is replaced by a ring $R$. In this regard, I was looking for simple examples for failure.

  • $\begingroup$ Are you happy with examples where $|G|=1$? $\endgroup$ – Lord Shark the Unknown Oct 13 '18 at 4:55
  • $\begingroup$ No; I am sorry for not mentioning this explicitly in the question. Should I edit? $\endgroup$ – Beginner Oct 13 '18 at 4:56
  • 3
    $\begingroup$ Well, if you have examples where $|G|=1$, you can turn them into examples over any nontrivial group $G$ by just letting $G$ act trivially. $\endgroup$ – Eric Wofsey Oct 13 '18 at 7:43

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