For a sufficient large field $k$ with characteristic 2, $S\_3$ and $D\_{10}$ both do have the property that the trivial module over their group algebra has nontrivial self-extension. Puig's work on nilpotent blocks shows that being p-nilpotent for the considered characteristic is sufficient to guarantee the trivial module has nontrivial self-extension and $S\_3$ and $D\_{10}$ are 2-nilpotent. But if we change the characteristic of $k$ from 2 to 3,5 respectively, direct computation shows that the trivial module over the group algebras $kS\_3$ and $kD\_{10}$ only have trivial self-extension respectively. However we know that they are all solvable, thus being solvable is not a sufficient condition on $G$ for the trivial module over $kG$ has nontrivial self-extension. Fixed a algebraically closed field $k$ of characteristic $p$, is there any other criterion for the trivial module over the group algebra $kG$ has nontrivial self-extension except for that $G$ is $p$-nilpotent?



Nontrivial self-extension of trivial $kG$-module
$\mathrm{Ext}^1\_{kG}(k,k)$ classifies these extensions, and it's not too hard to show that $\mathrm{Ext}^1\_{kG}(k,k) \cong H^1(G;k) \cong \hom(G, k)\cong \bigoplus\_{i\in I}\hom(G^{ab},\mathbb F\_p)$ where $|I| = \dim\_{\mathbb F\_p}(k)$.


In particular, this is nonzero if and only if $\hom(G^{ab},\mathbb F\_p)\neq 0$, if and only if $G^{ab}$ has a nontrivial $p$-group quotient. This is equivalent to $G$ having a nontrivial $p$-group quotient, which is weaker than being $p$-nilpotent.


