Schur-triviality is not subgroup-closed

This article gives the statement, and possibly proof, of a group property (i.e., Schur-trivial group) not satisfying a group metaproperty (i.e., subgroup-closed group property).
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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.

Proof

Example of semidihedral group

We take the following:

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

Example of general linear group

We could take:

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