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Let $\chi$ be the character of a representation of a simple group $G$ and let $g\in G$. If $g$ has order two and $G\neq C_2$ then show that $\chi(g)\equiv \chi (e)$ modulo 4. The hint I get is to consider the action of the determinant of $g$ acting on $V$.

$g$ has order two so the eigenvalues of $g$ must all be $\pm 1$. So I think somehow I need to show $\det (g)=1$. But how?

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2 Answers 2

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The elements $a$ with $\det a=1$ form a normal subgroup. If this is $G$, we are done. If it is $\{e\}$, then $\det g=-1$. But the elements $a$ with $\det a=\pm1$ also form a normal subgroup. This must be $G$; but unless $G=C_2$ it's impossible that one element has determinant $1$ and all others $-1$.

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Suppose $\rho:G\to\operatorname{End}(V)$ is your representation. Then the determinant gives a group homomorphism $\delta:g\in G\mapsto\det\rho(g)\in\mathbb C^\times$. Since $G$ is simple, the map $\delta$ must be or injective or trivial. The first case can only happen if $G$ is cyclic of prime order; if it has elements of order $2$, then it must be $C_2$ and it isn't. It follows that $\delta$ is trivial. In particular, $\det(g)=\delta(g)=1$.

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