most recent 30 from http://mathoverflow.net 2013-05-20T08:59:39Z http://mathoverflow.net/feeds/question/114484 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114484/subgroups-of-a-p-solvable-group-and-complete-reducibility sife 2012-11-26T04:04:41Z 2012-11-26T09:09:11Z

<p>1.</p> <p>Let $G$ be a $p$-solvable group and $V$ be a finite dimensional faithful $kG$-module, where the characteristic of $k$ is $p$. But $V$ is not a semisimple $kG$-module. For every $n\geq 0$, we define $End_{k}^{0}(V)=V$ and $End_{k}^{n}(V)=End_{k}(End_{k}^{n-1}(V))$ when $n>0$.</p> <p>Let $L$ be a subgroup of $G$ such that $End_{k}^{n}(V)$ is a semisimple $kL$-module for every $n\geq 0$.</p> <p>Is $L$ a $p'$-subgroup?</p> <p>2.</p> <p>Let $V$ be a 2-dimensional vector space over a field $k$ of characteristic $p$ and $G$ be a $p$-solvable subgroup of $GL(V)$, where $p$ is a prime number larger than than 5. But $V$ is not a semisimple $kG$-module.</p> <p>Let $L$ be a subgroup of $G$ such that $V$ is a semisimple $kL$-module.</p> <p>Is $L$ a $p'$-subgroup?</p>

http://mathoverflow.net/questions/114484/subgroups-of-a-p-solvable-group-and-complete-reducibility/114505#114505 Peter Mueller 2012-11-26T09:09:11Z 2012-11-26T09:09:11Z

<p>The answer to the second question is yes: As $G$ is reducible, this is true for $L$ even more. As $V$ is a semisimple $L$-module, $V$ is a sum of two $1$-dimensional $L$-modules. So $L$ is conjugate to a group of diagonal matrices. But elements of finite multiplicative order in $k$ have $p'$-order. </p>