Schur-triviality is not quotient-closed

Statement
It is possible to have a Schur-trivial group $$G$$ (i.e., the Schur multiplier of $$G$$ is the trivial group) and a normal subgroup $$H$$ of $$G$$ such that the quotient group $$G/H$$ is not Schur-trivial.

Example of the quaternion group
This is the smallest example:


 * $$G$$ is the quaternion group, which is a Schur-trivial group.
 * $$H$$ is the center of quaternion group, which is of order two, and is isomorphic to cyclic group:Z2.
 * $$G/H$$ is isomorphic to the Klein four-group, which is not Schur-trivial.

Perfect group examples
We can construct many examples this way: let $$G$$ be the universal central extension of some perfect group $$M$$ that is not Schur-trivial. Then, $$G$$ is a Schur-trivial group, and $$M$$ is the quotient of $$G$$ by some central subgroup, and is not Schur-trivial.