# Relation between the order of an element of a group and their character in a simple group

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.

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.