 Solution

Use the continuous determinant mapIt induces a homomorphism on fundamental groups. Under the standard identificationthe class of $t\mapsto e^{2\pi it}$ corresponds to its winding number $1$. Sincewe obtainFor $k\ne 0$ this is not the identity, so $[\gamma {]}^k$ could not have been the identity in ${\pi }_1(U(2),I)$.