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