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Yes. In fact, $V$ has an infinite dimensional semisimple quotient, which is decomposable since any simple $\mathbb{F}_pG$-module is a quotient of $\mathbb{F}_pG$ and so is finite-dimensional. Let $V' …

The condition on Brauer characters is not sufficient. Let $G$ be a $p$-group, $\pi$ any nontrivial representation over $\mathbb{F}_p$, and $\sigma$ the trivial representation over $\mathbb{Q}_p$ of t …

Representations of $B$ (or at least an equivalent category) are studied in the literature under the name of "cohomological Mackey functors". Theorem 1.1 of Bouc, Serge; Stancu, Radu; Webb, Peter, On t …

Are you sure you mean $p^k$, and not something like $p^{k^2}$? If you look at indecomposable $d$-dimensional representations of the free algebra $\mathbb{F}_p\langle x,y\rangle$, the number grows fas …

I presume you want $\mathcal{O}$ to have characteristic zero and $F$ to have characteristic $p$. Consider the case where $p=2$, $G=C_2$ and $P=G$. If $\mathcal{O}$ is the trivial $\mathcal{O}G$-module …

To answer the first question, there is a larger Krull-Schmidt category than $\text{mod}\,A$ unless $A$ has finite representation type. Every indecomposable pure-injective module has local endomorphism …

At least if the field is algebraically closed (or sufficiently large), a finite group has a normal $p$-complement if and only if its principal block is local. For example, this is Corollary 6.13 of Na …

Example 5.10 of Towers, Matthew, Endomorphism algebras of transitive permutation modules for $p$-groups., Arch. Math. 92, No. 3, 215-227 (2009) (whose author you might know) gives a positive answer to …

The two notions are the same. Clearly the first implies the second. Assume that $\sigma^{(m)}$ is isomorphic to $\sigma$. So there is some $a\in GL_n(k)$ such that $a\sigma^{(m)}(g)a^{-1}=\sigma(g)$ f …

Take $p=2$, $k$ algebraically closed, $A=kC_3$, and $H=C_2$ acting non-trivially on $C_3$. Then $A$ has three simple modules, but $A\# kH$ is the group algebra $kS_3$, which has two simple modules.

For question 2 and $S_n$, suppose $\text{char}(K)=p$ and $KS_n$ is a Nakayama algebra. Then $KS_n$ must have finite representation type, and so $S_n$ must have cyclic Sylow $p$-subgroups, which means …

I'll give three answers, which basically say: (A) it doesn't matter, (B) it's not true, and (C) here's (a sketch of) a proof. But before that, there are a couple of relevant conditions in Linckelmann' …

For (commutative) group algebras $A=kG$ the answer is no. Suppose $k$ has characteristic $p$, and $G=P\times H$, where $P$ is a Sylow $p$-subgroup. Then $kG\cong kP\otimes_kkH$, which, since $kH$ is …