Schur multiplier of Z-group is trivial

Statement
Suppose $$G$$ is a Z-group, i.e., $$G$$ is a finite group such that every Sylow subgroup of $$G$$ is a cyclic group (and in particular, a finite cyclic group). Then, $$G$$ is a Schur-trivial group: the Schur multiplier of $$G$$ is the trivial group.

Facts used

 * 1) uses::Cyclic implies Schur-trivial
 * 2) uses::All Sylow subgroups are Schur-trivial implies Schur-trivial, which in turn is a corollary of the fact that uses::finite group generated by Schur-trivial subgroups of relatively prime indices is Schur-trivial

Proof
The proof follows directly from facts (1) and (2).