Does $GL_5(\mathbb{R})$ has subgroup of index $2$?
My answer - Yes.
Let's say that $\tau$ is a function, which is the sign of the determinant of the matrices in $GL_5(\mathbb{R})$ ( easy to see that it is homomorphism).
Then we have $Im(\tau)=\{1,-1\}$, and by the first isomorphism theorem, we get that the index of $ker(\tau)$ is $2$.
Is it correct? Did I miss something?
You are correct, but your way of explaining what you proved is a bit clumsy. Instead of writing “Let's say that $\tau$ is a function, which is the sign of the determinant”, you could have said that, if $M\in GL_5(\mathbb{R})$, you define $\tau(M)=\operatorname{sgn}(\det M)$