Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G$ be a group, $F$ a field, and $V$ be an $F[G]$ module (equivalently $F$-representation of $G$). The following definition is well-known.

Definition 1. We say that $V$ is irreducible (or simple $F[G]$-module) if there is no proper non-zero subspace (submodule) $W$ of $V$ such that $g.W\subseteq W$ $\forall g\in G$.

Question: Can we reformulate this definition in the following way?

Definition 2. We say that $V$ is irreducible (or simple $F[G]$-module) if for some $0\neq v\in V$, $\langle g.v \,\colon g\in G\rangle =V$.

share|improve this question
    
The usual definition of irreducible also requires the module $V$ to be nonzero. –  Marc van Leeuwen Dec 1 '12 at 10:47
add comment

2 Answers

up vote 8 down vote accepted

No. $F[G]$ is an $F[G]$-module, apparently generated by 1, but is reducible, since it contains the submodule $\langle \sum_g g \rangle$.

share|improve this answer
    
(and that subspace is a submodule) –  mt_ Dec 1 '12 at 10:12
    
@mt_, Thanks! ${}$ –  user27126 Dec 1 '12 at 10:38
add comment

Sanchez is perfectly correct. In fact what you define is a cyclic module. Notice that if there is such a $v,$ then $a \to a.v$ for all $a \in FG$ is a module epimorphism, so the module $V$ is an epimorphic image of the regular module $FG.$ Conversely, if there is an epimorphism $\phi: FG \to V$ of $FG$ modules, then $\phi(1_{G})$ generates $V$ as $FG$-module. It s possible to go further, and to show that (for any field $F$) an $FG$-module $M$ is a cyclic module if and only if $M/{\rm rad}(M)$ is a direct sum of pairwise non-isomorphic $FG$-modules, where ${\rm rad}(M)$ is the smallest submodule of $M$ such that $M/{\rm rad}(M)$ is completely reducible.

share|improve this answer
add comment

Your Answer

 
discard

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.