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Let $F=\mathbb{F}_q$ be the field of $q$ elements and $G$ be a finite group. I'm trying to show that for an irreducible $FG$-module $V$, we have $\mathrm{End}_{FG}(V)=F \cdot 1$, i.e. that $F$ is a splitting field for $FG$.

My attempts thus far at this problem have involved trying to show that $V$ must be absolutely irreducible, as I believe this is equivalent to the statement I am trying to prove. This is where I am at a loss. I have tried to show that $V \otimes_F \mathbb{F}_{p^{n!}}$ is irreducible for all $n$ (as $\mathbb{F}_{p^{\infty}}$ is the algebraic closure of $\mathbb{F}_q$), though I'm not too sure exactly how to do this. I'm feeling a little bit out of my depth here so any help/pointers would be greatly appreciated!

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  • $\begingroup$ I really am not sure what do you mean by "splitting field of a ring (or algebra) group"...do you have some link/quote for this definition, please? $\endgroup$ – DonAntonio Apr 28 '13 at 12:22
  • $\begingroup$ The definition of "K is a splitting field of an algebra R" is simply taken to mean that for each irreducible R-module V (where R is an algebra over a field K), we have End_R(V) = K 1. In other words, it is simply as defined in the first paragraph. Sorry if I was not clear in this. I have taken this from the textbook "Modular Representations of Finite Groups" by Puttaswamaiah & Dixon (pg. 11, definition). I believe a preview for this is on Google books. $\endgroup$ – Iteraf Apr 28 '13 at 13:05
  • $\begingroup$ See this question for more counterexamples. Get into the habit of testing claims like this with cyclic groups first. Here they do produce irreducible reps that are not absolutely irreducible! Observe that we should except all the absolutely irreducible reps of a finite cyclic group to be 1-dimensional as they do over the complex numbers (at least when the order of the group is coprime to $p$ and Maschke applies). $\endgroup$ – Jyrki Lahtonen Apr 28 '13 at 13:54
  • $\begingroup$ ^expect ${}{}{}$ $\endgroup$ – Jyrki Lahtonen Apr 28 '13 at 17:17
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The claim is false as stated. For a small example let $G$ be the cyclic group of order three generated by the element $g$, and let $q=2, F=\mathbb{F}_2$. Let $$ V=\{a_01+a_1g+a_2g^2\in FG\mid a_0+a_1+a_2=0\}. $$ It is easy to see that $V$ is irreducible. Yet multiplications by fixed elements of $G$ obviously give distinct non-zero endomorphisms of $V$ so there are four endomorphisms of $V$.

You correctly judged that this has a lot to do with absolute irreducibility. This module $V$ splits into a sum of two 1-dimensional representations over $E=\mathbb{F}_4$. Let $\omega$ be a primitive cubic root of unity in $E$. Then as modules over $EG$ we get $$ E\otimes_FV=E(1+\omega g+\omega^2g^2)\oplus E(1+\omega^2g+\omega g^2). $$ It is also easy to see that $\operatorname{End}_{FG}(V)\cong E$ as $F$-algebras.

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  • $\begingroup$ I think that if you add the assumption $|G|\mid q-1$ then you get absolute irreducibility. That will do it for cyclic subgroups of $G$, and IIRC that suffices?? $\endgroup$ – Jyrki Lahtonen Apr 28 '13 at 15:19

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