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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This article gives a proof/explanation of the equivalence of multiple definitions for the term Z-group
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Statement

Suppose $G$ is a Z-group. In other words, $G$ 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 (?) 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.

Facts used

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