subgroups of a \$p\$-solvable group and complete reducibility - MathOverflow 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 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 Answer by Peter Mueller for subgroups of a \$p\$-solvable group and complete reducibility 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>