= Solution
The <internal direct product theorem> states that if $H,K\trianglelefteq G$, $H\cap K=\{1\}$, and $HK=G$, then $G\cong H\times K$. Indeed, for $h\in H,k\in K$, the commutator $[h,k]$ lies in both $H$ and $K$, so it is trivial. The map $(h,k)\mapsto hk$ is consequently a homomorphism; the intersection condition makes it injective and $HK=G$ makes it surjective. Conversely, the two factors in a direct product satisfy these conditions.
In odd dimension, $-I$ is central and has determinant $-1$. Every $Q\in O(3)$ has a unique expression $Q=(-I)^\varepsilon R$ with $R\in SO(3)$, so
$$
\boxed{O(3)\cong SO(3)\times C_2.}
$$
By contrast, $SO(2)\times C_2$ is abelian, while reflections in $O(2)$ conjugate a rotation to its inverse. Thus \b[$O(2)\not\cong SO(2)\times C_2$].
The center of the <unitary group> consists exactly of scalar unitary matrices:
$$
\boxed{Z(U(2))=\{zI:|z|=1\}.}
$$
The determinant gives a surjective homomorphism
$$
\det:U(2)\longrightarrow\mathbb T
$$
whose kernel is the <special unitary group> $SU(2)$.
Nevertheless $U(2)$ is not isomorphic to $SU(2)\times\mathbb T$. The center of $U(2)$ has exactly one nonidentity element of order two, namely $-I$. The center of the proposed product is $\{\pm I\}\times\mathbb T$, which has three nonidentity elements of order two. Since an isomorphism preserves the center and element orders,
$$
\boxed{U(2)\not\cong SU(2)\times\mathbb T.}
$$
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