# Simple module question [closed]

Let $p$ be a prime number, $G$ be a non-trivial finite $p$-group, $F$ be a field of characteristic $p$, and $FG$ be the group ring.

I want to prove that the unique simple $FG$-module (up to isomorphism) given by $FG/rad(FG)$ is neither projective nor injective.

## closed as off-topic by mathematics2x2life, Rhys Steele, user21820, Did, A. GoodierMar 30 '18 at 10:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – mathematics2x2life, Rhys Steele, user21820, Did, A. Goodier
If this question can be reworded to fit the rules in the help center, please edit the question.

• Please do not delete questions which have answers. – Mariano Suárez-Álvarez Mar 29 '18 at 17:47
• @MarianoSuárez-Álvarez I agree with your admonestation, of course, but, before that, why choose to answer any of the questions with no context from this OP? For a relevant comment, please see here. – Did Mar 30 '18 at 10:05
• @Did, when I answered this I didi not look at the history of the user. I never do. He is obviously cheating homework or something like that, which is sad, but who ever gave him/her the exercise has good taste. (The need for context is flexible: i honestly do not think that in this particular case there is much 'context' to be added. It is the sort of thing that unless one has an idea one may well have no idea of how to attack) – Mariano Suárez-Álvarez Mar 30 '18 at 18:35
• In the name of all that is holy why would anyone vote to delete a question which has an answer?!?! Yes: the question could have been better in many ways, but that ship has sailed. Whatever ideas one has that tell one that having an answered question is a worse option that not having it, independently of the question, are simply absurd. Vote to close, whatever: but deletion? Really?! – Mariano Suárez-Álvarez Apr 4 '18 at 1:48
• I can see the argument for not deleting this question, but what's the rationale behind the votes to reopen? What purpose is served by reopening it? – G Tony Jacobs May 12 '18 at 22:23

$\require{AMScd}$Since the injectives are precisely the projectives, it is enough to show that your module is not projective. Suppose then that $FG/\operatorname{rad} FG$ is projective. This module is he trivial one-dimensional module.

Suppose now that $f:M\to N$ is a surjection of $G$-modules. Let $\hom(M,N)$ denote the vector space of all linear maps $M\to N$, not necessarily $G$-linear, and similarly for $\hom(N,N)$. The map $$f_*:h\in\hom(N,M)\mapsto f\circ h\in\hom(N,N)$$ is surjective. If we regard $\hom(N,M)$ and $\hom(N,N)$ as $G$-modules in the usual way, the map $f_*$ is $G$-linear. Now consider the linear map $q:K\to\hom(N,N)$ which maps $1$ to the identity of $N$: this map is in fact $G$-linear, and therefore we have a diagram with exact row of $G$-modules and $G$-linear maps

                K
|
| q
|
V
hom(N,M) ———> hom(N,N) ———>  0
f_*


Since $K$ is projective, there exists a $G$-linear map $Q:K\to \hom(N,M)$ such that $f_*\circ Q=q$.

You should have no trouble showing that $g=Q(1)$, which is a linear map $N\to M$, is in fact $G$-linear, and that $f\circ g=1_N$. This means that the $G$-linear map $f:M\to N$ has a left inverse. We thus see that $N$ is projective.

This shows that if $K$ is projective, then all modules are projective and, therefore, that $KG$ is semisimple. But it isn't

• Notice that this argument shows in fact that if the trivial module over $FG$ is projective, then $FG$ is semisimple for all $F$ and finite $G$. This is a form of Maschke's theorem. – Mariano Suárez-Álvarez Mar 29 '18 at 8:32
• I don't understand why ''Since the injectives are precisely the projectives" can you please explain more? thanks – aymano Mar 29 '18 at 8:51
• Because group algebras are Frobenius algebras and therefore they are self-injective. – Mariano Suárez-Álvarez Mar 29 '18 at 8:54
• do you mean that we have alway the implication : if $M$ is injective module then it si a projective module? – aymano Mar 29 '18 at 8:56