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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 $0\neq v\in V$, $\langle g.v \,\colon g\in G\rangle =V$.

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The usual definition of irreducible also requires the module $V$ to be nonzero. – Marc van Leeuwen Dec 1 '12 at 10:47
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 $\langle \sum_g g \rangle$.

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(and that subspace is a submodule) – m_t_ Dec 1 '12 at 10:12
@mt_, Thanks! ${}$ – user27126 Dec 1 '12 at 10:38

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.

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