Let X be a real orientable compact differentiable manifold. Is the (co)homology of X generated by the fundamental classes of oriented subvarieties? And if not, what is known about the subgroup generated?
Rene Thom answered this in section II of "Quelques propriétés globales des variétés différentiables." Every class $x$ in $H_r(X; \mathbb Z)$ has some integral multiple $nx$ which is the fundamental class of a submanifold, so the homology is at least rationally generated by these fundamental classes.
Section II.11 works out some specific cases: for example, every homology class of a manifold of dimension at most 8 is realizable this way, but this is not true for higher dimensional manifolds and the answer in general has to do with Steenrod operations.
This is a reply to Alon's comment, but it's too long to be a comment and is probably interesting enough to be an answer.
Here's an example Thom gives of a homology class that is not realized by a submanifold: let $X=S^7/\mathbb Z_3$, with $\mathbb Z_3$ acting by rotations, and $Y=X \times X$.
Then $H^1(X;\mathbb Z_3)=H^2(X;\mathbb Z_3)=\mathbb Z_3$ (and they are related by a Bockstein); let $u$ generate $H^1$ and $v=\beta u$ be the corresponding generator of $H^2$. Then it can be shown that the class $u \otimes vu^2 - v \otimes u^3 \in H^7(Y;Z_3)$ is actually integral (i.e., in $H^7(Y;Z)$), and its Poincare dual in $H_7$ cannot be realized by a submanifold (in fact, it can't be realized by any map from a closed manifold to $Y$, which need not be the inclusion of a submanifold). This is a natural example to consider because the first obstruction to classes being realized by submanifolds comes from a mod 3 Steenrod operation, and these are easy to compute on $Y$ because $X$ is the 7-skeleton of a $K(\mathbb Z_3,1)$. Note that the class in question is 3-torsion, so trivially 3 times it is realized by a submanifold.