All Sylow subgroups are Schur-trivial implies Schur-trivial

Definition
Suppose $$G$$ is a finite group such that for every prime number $$p$$, the $$p$$-Sylow subgroup of $$G$$ is a Schur-trivial group. Then, $$G$$ is also a Schur-trivial group.

Note that even though a group may have more than one $$p$$-Sylow subgroup for any fixed prime $$p$$, Sylow implies order-conjugate, hence isomorphic, so one is Schur-trivial if and only if all the others are.

Related facts

 * Schur multiplier of cyclic group is trivial

Facts used

 * 1) uses::Finite group generated by Schur-trivial subgroups of relatively prime indices is Schur-trivial

Proof
The proof follows from Fact (1), along with the observation that a collection containing one $$p$$-Sylow subgroup for every prime divisor $$p$$ of the order of $$G$$ satisfies the conditions of Fact (1). Note that the "relatively prime" of Fact (1) refers to the collective gcd of all the indices, and is weaker than saying "pairwise relatively prime."