Schur multiplier of Z-group is trivial

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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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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


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