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Let $\chi$ be the representation of a finite group $G$. Let $g \in G$ be an element of order 2. If $G$ is a simple group but not cyclic of order 2, prove that $\chi(g) \equiv \chi(1) \mod 4$. Proof that $\chi(g) \equiv \chi(1) \mod 2$ in any finite group can be found here: Relation between the order of an element of a group and their character but I'm having trouble connecting the two. Any help would be appreciated.

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Let $\rho$ be a representation affording $\chi$. Then the eigenvalues of $\rho(g)$ are $\pm 1$. Suppose $m$ of them are $1$ and $n$ are $-1$. So $\chi(g) = m-n$ and $\chi(1) = m+n$.

If $\chi(g) \equiv 2 \bmod \chi(1)$, then $n$ is odd, and so $\det(\rho(g)) = -1$. But since the map $\tau:G \to {\mathbb C}^*$ defined by $\tau(g) = \det(\chi(g))$ is a group homomorphism with image a finite cyclic group, this would imply that $G$ had a normal subgroup of index $2$, contrary to hypothesis.

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