Schur-triviality is not subgroup-closed

Statement
It is possible to have a Schur-trivial group $$G$$ and a subgroup $$H$$ of $$G$$ that is not Schur-trivial.

Related facts

 * For finite groups, being a group for which every subgroup is Schur-trivial is equivalent to being a finite group with periodic cohomology, which in turn is equivalent to every abelian subgroup being cyclic.

Example of semidihedral group
We take the following:


 * $$G$$ is particular example::semidihedral group:SD16, which is a Schur-trivial group.
 * $$H$$ is the subgroup particular example::D8 in SD16 inside $$G$$.
 * $$H$$ is abstractly isomorphic to particular example::dihedral group:D8, which is not Schur-trivial.

Example of general linear group
We could take:


 * $$G$$ is particular example::general linear group:GL(2,3), which is a Schur-trivial group of order 48.
 * $$H$$ is the subgroup particular example::D8 in GL(2,3), obtained by using the faithful irreducible representation of dihedral group:D8 over field:F3.
 * $$H$$ is isomorphic to particular example::dihedral group:D8, which is not Schur-trivial.