Every Sylow subgroup is cyclic implies metacyclic

This article describes a result about finite groups where the isomorphism types of the Sylow subgroup (?)s implies certain properties of the whole group. Note that all $p$-Sylow subgroups for a given prime $p$ are conjugate and hence are isomorphic, so the statement makes sense.
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Statement

Suppose $G$ is a finite group with the property that every Sylow subgroup (?) of $G$ is cyclic, i.e., is a Cyclic Sylow subgroup (?). Then, $G$ is a Metacyclic group (?): it has a Cyclic normal subgroup (?) (in fact, a cyclic Normal Hall subgroup (?)) such that the quotient is also a cyclic group.

Related facts

Applications

• Square-free implies solvability-forcing

References

Textbook references

• The Theory of Groups by Marshall Hall, Jr., Page 146, Theorem 9.4.3, (Proof uses counting arguments and is about two pages long.)^{More info}