First group cohomology and composition factors

Question

Let $G$ be a finite group. Let $k$ be a field ($\text{char}(k)=p>0$). Let $P(k)$ be the projective cover of $k$. Assume that for any nontrivial simple $kG$-module $M$ we have $H^1(G,M)=0$. Does it mean that all composition factors of $P(k)$ are trivial?

Answer 1

Yes. $H^1(G,M) = \operatorname{Ext}_{kG}^1(k,M)$ describes how many ways $M$ can be a submodule of $N$ with $N/M \cong k$. $P(k)$ can only have composition factors that are “connected” to $k$ by paths of nonzero $\operatorname{Ext}_{kG}^1$. We prove this by induction on the Loewy length, that is, by considering the semisimple modules $J^n P(k) / J^{n+1} P(k)$ and their extensions $J^{n-1} P(k) / J^{n+1} P(k)$ where $J$ is the Jacobson radical of $kG$.

Lemma: $P(k) / J^2 P(k)$ only has trivial composition factors.

Proof: Let $Q$ be a maximal submodule of $J P(k)$ setting $M=J P(k)/Q$ to be the simple quotient module. Consider the short exact sequence $0 \to M \to P(k)/Q \to k \to 0$. Assume by way of contradiction that $M$ is not the trivial (central/principal) module $k$. Since $\operatorname{Ext}_{kG}^1(k,M) = 0$, the sequence splits. However, $J P(k)/Q$ is superfluous in $P(k)/Q$, so never has a supplement, much less a complement. This contradiction shows $M=k$. $\square$

Proposition: $P(k)$ only has trivial composition factors.

Proof: Suppose $P(k)/J^n P(k)$ only has composition factors isomorphic to $k$, and let $R$ be a maximal submodule of $J^n P(k)$. Set $M=J^n P(k)/R$ to be the simple quotient, and assume by way of contradiction that $M$ is not the trivial (central/principal) module $k$. Consider $$0 \to M \to J^{n-1} P(k)/R \to J^{n-1} P(k)/ J^n P(k) \to 0$$ and notice $J^{n-1} P(k)/ J^n P(k) \cong k^m$ for some positive integer $m$. Hence $$\operatorname{Ext}_{kG}^1(J^{n-1} P(k)/ J^n P(k),M) = \operatorname{Ext}_{kG}^1(k^m, M) = \operatorname{Ext}_{kG}^1(k,M)^m = 0^m = 0$$ so the sequence splits. However, $M = J^n P(k) / R = J ( J^{n-1} P(k)/R )$ is superfluous, so not a direct summand. This contradiction shows that in fact $M$ must be isomorphic to $k$. Since $J^n P(k) / J^{n+1} P(k)$ is isomorphic to the direct sum of such $M$, the induction proceeds to show that $J^n P(k) / J^{n+1} P(k)$ and hence $P(k) / J^{n+1} P(k)$ only has trivial composition factors. Since $P(k)$ has finite Loewy length, $P(k)$ itself only has trivial composition factors. $\square$

The more intuitive method view $P(k)$ made up of layers. The top layer is a single copy of $k$, and the second layer is proven in the lemma to only have copies of $k$ again. What can happen in the third layer? Well the third layer of $P(k)$ is actually the second layer of $J (k)$ and since $J P(k)$ is a homomorphic image of $P(k)^m$, its second layer also only has copies of $k$. Thinking of this in terms of the original copy of $P(k)$, we see the third layer of $P(k)$ also only has trivial composition factors. In other words the $n$th layer of $P(k)$ is the second layer of $J^{n-2} P(k)$, and if $J^{n-2} P(k)$ only has trivial top composition factors, then it is a homomorphic image of $P(k)^m$, etc.

This kind of thing is uncommon in finite groups. I believe this is only the case when the principal block contains only one simple module. I think this is the case when $G/O_{p'}(G)$ is a $p$-group, but I don't have my references around to double check.