# one dimensional representation is irreducible

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 $g\in G$.

-

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 $V=\langle v_0 \rangle\subset W$? –  palio Nov 18 '11 at 14:51
@ 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. –  Ben Blum-Smith Nov 18 '11 at 14:55