Schur multiplier of Z-group is trivial

This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., Z-group) must also satisfy the second group property (i.e., Schur-trivial group)
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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. Cyclic implies Schur-trivial
2. All Sylow subgroups are Schur-trivial implies Schur-trivial, which in turn is a corollary of the fact that finite group generated by Schur-trivial subgroups of relatively prime indices is Schur-trivial

Proof

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