# 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.

## 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
• Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page 258, Theorem 6.2, Chapter 7 (Fusion, transfer and p-factor groups), Section 7.6 (elementary applications), More info