MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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 $n>0$.

Let $L$ be a subgroup of $G$ such that $End_{k}^{n}(V)$ is a semisimple $kL$-module for every $n\geq 0$.

Is $L$ a $p'$-subgroup?


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.

Let $L$ be a subgroup of $G$ such that $V$ is a semisimple $kL$-module.

Is $L$ a $p'$-subgroup?

share|cite|improve this question

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.