$S^0$, $S^1$, and $S^3$ have well-known smooth group structures. These can be obtained from well-known bilinear multiplications in $\mathbb R$, $\mathbb R^2$, and $\mathbb R^4$. $S^7$ does not have a smooth group structure; this fact can be obtained from the theory of Lie groups and their Lie algebras. Stronger statement: $S^7$ does not have a continuous group structure; this can be deduced from the previous fact using the deep solution of a Hilbert problem. But $S^7$ does have an $H$-space structure, by which I mean a continuous multiplication law with a two-sided identity. This can be obtained from a (non-associative) bilinear multiplication on $\mathbb R^8$, so it can even be chosen in such a way that multiplication is smooth and multiplication by a fixed element on either side has a smooth inverse.

Outside these dimensions, $S^{n-1}$ has no $H$-space structure. For $n>1$ odd, this is easy using cup products in $S^{n-1}\times S^{n-1}$. For $n$ not a power of $2$, it can be done using Steenrod operations in mod $2$ cohomology. The general case uses $K$-theory.

A corollary is that real division algebras (even in a nonassociative sense) are impossible in dimensions other than $1$, $2$, $4$, and $8$. Another corollary is that no spheres other than those are parallelizable.

The question of how many linearly independent tangent vector fields the $(n-1)$-sphere admits was also settled (by Adams) using more subtle $K$-theory arguments.