Most likely here $Q$ is the generator of infinitesimal transformations that is a symmetry of your Hamiltonian. A finite transformation is then given by
$$g(q) = \exp(i q Q) = I + q Q + \mathcal O(q^2) $$
for some real number $q$. For example if $Q$ is the momentum operator, $g(q)$ is the translation operator (translation of distance $q$). If $Q$ is angular momentum along the $z$-axis, $g(q)$ is a rotation along $z$-axis of angle $q$, etc.
If the ground-state is invariant under this symmetry then we must have
$$ g(q)|0\rangle = |0\rangle,$$
for any $q$. This is true if
$$ Q|0\rangle = 0.$$
In more mathematical language, $Q$ belongs to the Lie Algebra of your symmetry group while $g(q)$ belongs to the Lie Group. The fact that the ground-state is invariant, means it transforms as the trivial representation of the Lie Group.