Finding a subgroup of index 2 using a representation Let $G$ be a group and $\rho \colon G \rightarrow GL(k, \mathbb{C})$ a representation. The question is to show that for an element of order 2, $g$ in $G$, $\chi(g) \equiv k$ (mod 4) given that there is no subgroup of index 2 in $G$. 
I have managed to reduce the problem to showing that if there is an element $A \in im(\rho)$ with $det(A) = -1$ then there must be a subgroup of index 2 in $G$.
If I let $\phi = det \circ \rho$ I can show that $\frac{G}{ker(\phi)}$ is abelian and of course $A \not \in ker(\phi)$.
Any hints on where to go from here? I'm hoping that $ker(\phi)$ is going to be the subgroup I'm looking for as I can't see anything else to test.
 A: Consider the determinant representation associated to $\rho$ (it is $1$-dimensional, and defined as $h \mapsto \det(\rho(h))$), and its character $\psi = \det(\chi)$. 
Let $h \in G$ be the element of order $2$. If you restrict $\rho$ to $\langle h \rangle$, it can (obviously) be put in diagonal form with only $1$ and $-1$ on the diagonal. Let $s$ be the number of $-1$, so $k-s$ is the number of $1$.
So you get that $\chi(h)=\operatorname{tr}(\rho(h))=(k-s)-s = k-2s$ (not really necessary, but interesting in its own right).
Also, you get $\psi(h)=(-1)^s$. If $s$ is odd, this means that $(-1)^s = -1 \in \operatorname{Im}(\psi)$. The image of $\psi$ is a subgroup of $\mathbb{C}^\times$, so it is cyclic, and since it contains $-1$ is has even order (it does not contain $0$, and if it contains $k$ it contains also $-k$ for all $k \in \mathbb{C}$). It has, therefore, a subgroup $K$ of index $2$.
Now you can define a subgroup $H=\{ g \in G \; | \; \psi(g) \in K\}$. This has index $2$ in $G$, a contradiction, hence $s$ cannot be odd (so it is even).
