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Let $G$ be a finite simple group, prove that $G$ does not have a non-trivial represention of degree $1$.

Remark: I intend to prove by contradiction. Let $V$ be a non-trivial $\mathbb{C}G$-module of $G$ with dimension $1$, let $H:=\{x\in G: vx=v, \forall v\in V \}$. Observe that $H$ is normal in $G$ and $H \neq G$ due to $V$ is not trivial, so $H=\{1\}$ which implies $V$ is faithful. But how does the existence of a faithful representation of degree $1$ lead to a contradiction?

Note: This problem arose when I tackle the other problem :"Let $G$ be a group of order 20. Prove that $G$ has a non-trivial representation of degree 1, and hence deduce that $G$ can not be simple." In which to prove $G$ is not simple I think it is equivalent to prove the above assertion.

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The problem should say "non-Abelian simple group". Hint: Show that a group with a faithful 1-dimensional representation of degree 1 is Abelian. – Geoff Robinson Oct 24 '12 at 6:24
I see now, thanks! – user31899 Oct 24 '12 at 6:45
up vote 3 down vote accepted

Let $G$ be a finite simple group. Let $f\colon G \rightarrow \mathbb{C}^\times$ be a non-trivial homomorphism. Since $G$ is simple, $Ker(f) = 1$. Hence $G$ is isomorphic to a finite subgroup of $\mathbb{C}^\times$. Hence $G$ is a finite cyclic group of a prime order.

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