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Mikhail Borovoi
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You can find the decomposition of $S^kV$ and $\Lambda^k V$ into a direct sum of irreducible representations in Table 5 in the book: Onishchik and Vinberg, Lie Groups and Algebraic Groups, Springer-Verlag 1990.

Instead of $U(n)$ you can consider its complexification ${\rm GL}(n)$. Then both $S^k V$ and $\Lambda^k V$ are irreducible.

For ${\rm SO}(n)$, $\Lambda^k V$ are irreducible, while $S^k V$ are not (indeed, $S^2 V$ contains the trivial representation).

For ${\rm Sp}(2n)$, $S^k V$ are irreducible, while $\Lambda^k V$ are not (indeed, $\Lambda^2 V$ contains the trivial representation).

Mikhail Borovoi
  • 14.2k
  • 2
  • 32
  • 72