Every Sylow subgroup is cyclic implies metacyclic

Statement
Suppose $$G$$ is a Z-group. In other words, $$G$$ is a finite group with the property that every fact about::Sylow subgroup of $$G$$ is cyclic, i.e., is a fact about::cyclic Sylow subgroup. Then, $$G$$ is a fact about::metacyclic group: it has a fact about::cyclic normal subgroup such that the quotient is also a cyclic group.

Further, we can choose the cyclic normal subgroup so that the quotient acts faithfully on it, and hence, the quotient is a subgroup of the automorphism group of the original cyclic normal subgroup.

Related facts

 * Classification of cyclicity-forcing numbers
 * Coprime automorphism group implies cyclic with order a cyclicity-forcing number

Applications

 * Square-free implies solvability-forcing

Facts used

 * 1) uses::Cyclic Sylow subgroup for least prime divisor has normal complement
 * 2) uses::Solvable implies Fitting subgroup is self-centralizing
 * 3) uses::Solvability is extension-closed: An extension of a solvable normal subgroup by a solvable quotient group is solvable.
 * 4) uses::Quotient group acts on abelian normal subgroup
 * 5) uses::Cyclic implies aut-abelian

Showing that the group is solvable
We induct on the order.

Given: A finite group $$G$$ of order $$n = p_1^{k_1}p_2^{k_2} \dots p_r^{k_r}$$ where $$p_i$$ are distinct primes. Every Sylow subgroup of $$G$$ is cyclic.

To prove: $$G$$ is solvable.

Proof: Without loss of generality, assume that $$p_1$$ is the least prime divisor of $$n$$. Let $$P_1$$ be a $$p_1$$-Sylow subgroup.


 * 1) Fact (1) tells us that $$P_1$$ has a normal complement $$N$$ in $$G$$.
 * 2) Every Sylow subgroup of $$N$$ is cyclic: $$N$$ has order $$p_2^{k_2} \dots p_r^{k_r}$$, so any Sylow subgroup of $$N$$ is Sylow in $$G$$, hence cyclic. Thus, we can apply the induction hypothesis, and obtain that $$N$$ is solvable.
 * 3) $$G$$ is solvable: $$G$$ has a solvable normal subgroup $$N$$ and a quotient $$G/N$$ that is a $$p_1$$-group, hence nilpotent (in fact, an application of the second isomorphism theorem yields that the quotient is isomorphic to $$P_1$$, which is cyclic). In particular, both $$N$$ and $$G/N$$ are solvable, so $$G$$ is solvable by fact (3).

Showing that the group is metacyclic
Given: A solvable finite group $$G$$ of order $$n = p_1^{k_1}p_2^{k_2} \dots p_r^{k_r}$$ where $$p_i$$ are distinct primes. Every Sylow subgroup of $$G$$ is cyclic.

To prove: $$G$$ is metacyclic, with a cyclic normal subgroup and a cyclic quotient acting faithfully on it by conjugation.

Proof: Let $$F(G)$$ be the Fitting subgroup of $$G$$.


 * 1) $$C_G(F(G)) \le F(G)$$: This follows from fact (2).
 * 2) $$F(G)$$ is cyclic: First note that $$F(G)$$ is nilpotent, so it is a direct product of its Sylow subgroups. Any Sylow subgroup of $$F(G)$$ is a $$p$$-subgroup of $$G$$, hence is contained in a $$p$$-Sylow subgroup. Thus, it is cyclic. This yields that $$F(G)$$ is cyclic.
 * 3) $$G/F(G)$$ acts faithfully on $$F(G)$$ by conjugation: Since $$F(G)$$ is cyclic, it is in particular abelian, so we get a well-defined action of the quotient group $$G/F(G)$$ on $$F(G)$$ (fact (4)). The kernel of this is $$C_G(F(G))/F(G)$$, but as we concluded in step (1), this is trivial. Thus, we have a faithful action of $$G/F(G)$$ on $$F(G)$$.
 * 4) $$G/F(G)$$ is cyclic: Since $$F(G)$$ is cyclic, its automorphism group is abelian, and from the previous step, $$G/F(G)$$ is isomorphic to a subgroup of $$\operatorname{Aut}(F(G))$$. Thus, $$G/F(G)$$ is abelian. On the other hand, all its Sylow subgroups are cyclic, since these are quotients of Sylow subgroups of $$G$$, which are cyclic. Thus, $$G/F(G)$$ is itself cyclic.

The conclusions of steps (2), (3), and (4) complete the proof.