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A comment on Torsten's answer: For any group $G$, and any subset $S \subset G$ it is true that $Z_G(Z_G(Z_G(S))) = Z_G(S)$. In particular the operation $Z_G$ is always an involution on its image.
You can also use an embedding in S^6 (a bit easier to construct). If nM = normal bundle, then H^2(S^6-nM) = Z by Mayer Vietoris. A geometric representative of the generating class will be a 4-manifold X with dX = M. More precisely, dX will be cobordant to a nonvanishing section of nM.
