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let $G$ be a finite group. let $V$ be an $F$-vector space. and $\rho:G\rightarrow GL(V)$ be a one dimensional representation. I don't see why it is automatically irreducible.

My guess: $V=\langle v_0 \rangle$ and $W$ a proper subspace of $V$ so $W\subset \langle v_0 \rangle$ an element in $W$ is $w=\alpha v_0$ for some $\alpha \in F$ so $\rho (g)(w)=\alpha \rho(g)(v_0)$ but why $W$ can't be invariant by $\rho(g)$ for any $g\in G$.

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A one dimensional vector space does not have a non-zero proper subspace!

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  • $\begingroup$ yes i was thinking about that but i could not see why if $W$ is a non zero subspace of a 1-dimensional vector space $V$ then we must have $V=\langle v_0 \rangle\subset W$? $\endgroup$ – palio Nov 18 '11 at 14:51
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    $\begingroup$ @palio, it would probably be a good idea for you to review some of your linear algebra before going on with group representations, then :) $\endgroup$ – Mariano Suárez-Álvarez Nov 18 '11 at 14:54
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    $\begingroup$ @ palio - one-dimensional means that every vector in $V$ has the form $cv_0$. If $W$ contains any nonzero vector $w_0$, then $w_0=cv_0$ for some nonzero $c$. But then because $W$ is closed under scalar multiplication, $v_0=c^{-1}w_0$ is also in it, and this means $V=\left<v_0\right>$ is in it. $\endgroup$ – Ben Blum-Smith Nov 18 '11 at 14:55
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Dimension of a representation is indeed the dimension of the vector space of representation. but dimension of a vector space is the cardinality of the basis. So a 1 dimensional vector space contains no non-zero proper subspace. Since the representation is one dimensional then it contains no proper subrepresentation. Therefore it is irreducible. Your question was quite trivial.

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