Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 3 / 6D / a / Solution

True. The determinant gives a group homomorphismFor an orthogonal matrix, $A^{T}A=I$ implies $(\det A{)}^2=1$, and a reflection has determinant $-1$, so the target has order two and the map is onto. Its kernel of a group homomorphism is exactlyThis is the defining special orthogonal group inside the orthogonal group.