Show that the unitary group $U(n)$ and the special unitary group $SU(n) \times S^1$ are diffeomorphic as manifolds, but not isomorphic as Lie groups when $n > 1$.
The above question is from a lecture note on differential topology. I can solve the first part of the question, but I am having difficulty showing that the two are not isomorphic as Lie groups when $n > 1$. I am given the hint to find a section $s : S^1 \to U(n)$ for the determinant function $\det : U(n) \to S^1$. The two obvious function I can think of is $a \mapsto \textrm{diag}(\sqrt[n]{a}, \dots, \sqrt[n]{a})$ and $a \mapsto \textrm{diag}(a,1,\dots,1)$, but I am not sure how considering either of the two smooth sections would help me seeing that the two Lie groups are nonisomorphic when $n > 1$.
I am not sure the hint is the best way to approach the question, but it may help to consider it as follows: if there is an isomorphism $\phi \colon \mathrm{U}(n) \to \mathrm{SU}(n)\times S^1$, say $\phi=(\phi_1,\phi_2)$ then by composing if necessary with an automorphism of $\mathrm{SU}(n)$, we may assume that $\phi_1 = \text{id}_{\mathrm{SU}(n)}$ and hence $\ker(\phi_2) = \mathrm{SU}(n)= \ker(\det)$ and thus we may identify $\phi_2$ with $\det$. That is, if such an isomorphism exists, then we may arrange that it has the form $\phi(u) = (\phi_1(u),\det(u))$ where $\phi_1\colon \mathrm{U}(n)\to \mathrm{SU}(n)$ and $\phi_{1|\mathrm{SU}(n)} = \text{id}_{\mathrm{SU}(n)}$.
But now if $\psi\colon \mathrm{SU}(n) \times S^1 \to \mathrm{U}(n)$ is the inverse of $\phi$ and we set $\psi_2(\lambda) = \psi(1,\lambda)$, then we have $$ \begin{split} \psi(s,\lambda) &= \psi((s,1)(1,\lambda)) = \psi(s,1)\psi(1,\lambda) = s\psi_2(\lambda)\\ &= \psi((1,\lambda)(s,1)) = \psi(1,\lambda)\psi(s,1) = \psi_2(\lambda)s \end{split} $$ and $\psi_2$ is a section of $\det$. But it is not hard to see that such a $\psi_2$ cannot exist, and hence that $\mathrm{U}(n)$ is not isomorphic to $\mathrm{SU}(n) \times S^1$.