# Examples of modules over a group ring

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.

• Are you happy with examples where $|G|=1$? – Lord Shark the Unknown Oct 13 '18 at 4:55
• No; I am sorry for not mentioning this explicitly in the question. Should I edit? – Beginner Oct 13 '18 at 4:56
• 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. – Eric Wofsey Oct 13 '18 at 7:43