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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?

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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 $\tau(M)=\operatorname{sgn}(\det M)$

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