# Projective but not free module over groupring

Let $G$ be a nontrivial finite group and consider the groupring $\Bbb QG$. My question is whether we can find a module over $\Bbb QG$ that is projective but not free?

By Maschke's Theorem, $\mathbb{Q}G$ is semisimple. In particular, every $\Bbb{Q}G$-module is projective. Thus, what you are looking for is any non-free module.

• Unfortunately, I'm not familiar with Maschke's Theorem. Is there an easy way to see that any $\Bbb QG$-module must be projective without appealing to Maschke's Theorem? Thanks. – iwriteonbananas Oct 21 '15 at 9:49
• On the top of my head, I cannot think of any other way, other than copying the proof of Maschke in this special case. However, I strongly encourage you to look at the proof of Maschke's Theorem. It is one of the fundamental result of representation theory and the proof is really short. – Nitrogen Oct 21 '15 at 14:29
• Thanks, will do. And what would be an example of a non-free $\Bbb QG$-module? – iwriteonbananas Oct 21 '15 at 14:45
• For any non-trivial group $G$, take $\Bbb{Q}$ with the trivial $G$-action. This module is not free because its $\Bbb{Q}$-dimension is $1$. If you want to show that it is projective by hand, you can use the maps $\Bbb{Q} \to \Bbb{Q}G:1 \mapsto \frac{1}{|G|}(\sum_G g)$ and $\Bbb{Q}G \to \Bbb{Q}:g \mapsto 1$ which show that $\Bbb{Q}$ (with the trivial action) is a direct summand of $\Bbb{Q}G$, and is hence projective. – Nitrogen Oct 21 '15 at 14:59
• My argument shows that $\Bbb{Q}$ is a direct summand because the map $\mathbb{Q} \to \mathbb{Q}G$ is injective and thus fits into a short exact sequence, which splits via the second map. If you want to use the lifting definition of projectivity, you have to be more subtle. Your map $f:\mathbb{Q} \to P$ lifts to a map $\tilde{f}:\mathbb{Q}\to R$ which sends $1$ to $\frac{1}{|G|}\sum_G g\cdot x$ where $x\in g^{-1}(f(1))$. I let you verify that $\tilde{f}$ is $\mathbb{Q}G$-linear (which is not necessarly the case for the $\tilde{f}$ you proposed) and that it lifts $f$. – Nitrogen Oct 21 '15 at 16:56

Any nontrivial direct summand $T$ of the right module $\Bbb Q G$, if such $T$ exists, is projective ( since summands of free modules are projective) but not free, since it can't have the dimension of a free module (which is clearly $|G|m$ where $m$ is a nonnegative integer. This argument applies to any finite group (other than $\{1\}$) and any field.

For fields of characteristic $0$ (like $\Bbb Q$) Maschke's theorem allows us to use any nontrivial right ideal at all, since they are all summands.

• Thanks for the answer. How do we know such a $T$ exists? Could you give me a concrete example? And why must the dimension of $T$ be something other than $|G|m$ with $m$ a nonnegative integer? – iwriteonbananas Oct 21 '15 at 10:49
• @iwriteonbananas How do we know such a $T$ exists? For fields not characteristic $0$ I guess I should mentioned that this existence is not guaranteed. However, in your case $\Bbb Q$ is characteristic $0$ and the group ring is semisimple. So your question "how do we know $\Bbb Q G$ has a nontrivial direct summand?" becomes "how do we know $\Bbb Q G$ has a nontrivial right ideal?" and the answer is "because it isn't a division ring." – rschwieb Oct 21 '15 at 11:07
• @iwriteonbananas why must the dimension of $T$ be something other than $|G|m$ with mm a nonnegative integer? Because $0<dim(T)<|G|$. – rschwieb Oct 21 '15 at 11:08
• Thank you. So we know that such a $T$ exists. Can we write down specifically an example of such? – iwriteonbananas Oct 21 '15 at 11:51
• @iwriteonbananas For any subgroup $H<G$, if you let $e=\frac{1}{|H|}\sum_{h\in H}h$, then $e\Bbb Q G$ is such a right ideal. – rschwieb Oct 21 '15 at 17:06