Trivial cobordism group in dimensions 1, 3, 7 related to H-space structures on the spheres in these dimensions? Is there a connection between the existence of H-space structures on $S^1$, $S^3$ and $S^7$ and the fact that every (closed) 1-manifold, 3-manifold and oriented 7-manifold is a boundary, or is this a pure numerical coincidence? 
[It's possible that my question amounts to groundless mysticism, but I had to wonder.]
On a related note: is it known which (oriented or not) cobordism groups are zero? 
 A: How about $n\ne 0,2,6$ instead of $n=1,3,7$? The Adams theorem on the vanishing of the stable Hopf invariant, which implies the non-existence of an H-structure on $S^n$ for $n\ne 1,3,7$, also implies the theorem of R. L. W. Brown and Liulevicius (and Mahowald) that if $M^n$ immerses in $\Bbb R^{n+1}$, then $M^n$ is a boundary or is cobordant to $\Bbb RP^0$, $\Bbb RP^2$ or $\Bbb RP^6$ (and these three do immerse in $\Bbb R^{n+1}$).
A: The cobordism groups can be calculated using the Adams spectral sequence, which is based on the homological algebra of modules over the Steenrod algebra.  This works most nicely for the unoriented cobordism groups, which form a polynomial algebra over $\mathbb{Z}/2$ with one generator in each degree not of the form $2^j-1$.  The powers of $2$ enter here as the degrees of the indecomposable generators $Sq^{2^j}$ in the Steenrod algebra.
The fact that $S^k$ can only have an H-space structure if $k$ has the form $2^j-1$ is also a straightforward consequence of the fact that the indecomposables in the Steenrod algebra occur in degrees $2^j$.  The fact that only $k=0$, $k=1$, $k=3$ and $k=7$ can appear was originally proved by Adams using the Adams spectral sequence and deeper analysis of the Steenrod algebra.  Thus, there are certainly some common themes between your two facts.  It would not surprise me if you could establish a more direct connection, but I do not immediately see one.
