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 -Sylow subgroups for a given prime are conjugate and hence are isomorphic, so the statement makes sense.
View other such results
This article gives a proof/explanation of the equivalence of multiple definitions for the term Z-group
View a complete list of pages giving proofs of equivalence of definitions
Suppose is a Z-group. In other words, is a finite group with the property that every Sylow subgroup (?) of is cyclic, i.e., is a Cyclic Sylow subgroup (?). Then, 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.
- Classification of cyclicity-forcing numbers
- Coprime automorphism group implies cyclic with order a cyclicity-forcing number
- Cyclic Sylow subgroup for least prime divisor has normal complement
- Solvable implies Fitting subgroup is self-centralizing
- Solvability is extension-closed: An extension of a solvable normal subgroup by a solvable quotient group is solvable.
- Quotient group acts on abelian normal subgroup
- Cyclic implies aut-abelian
Proof using advanced methods
Showing that the group is solvable
We induct on the order.
Given: A finite group of order where are distinct primes. Every Sylow subgroup of is cyclic.
To prove: is solvable.
Proof: Without loss of generality, assume that is the least prime divisor of . Let be a -Sylow subgroup.
- Fact (1) tells us that has a normal complement in .
- Every Sylow subgroup of is cyclic: has order , so any Sylow subgroup of is Sylow in , hence cyclic. Thus, we can apply the induction hypothesis, and obtain that is solvable.
- is solvable: has a solvable normal subgroup and a quotient that is a -group, hence nilpotent (in fact, an application of the second isomorphism theorem yields that the quotient is isomorphic to , which is cyclic). In particular, both and are solvable, so is solvable by fact (3).
Showing that the group is metacyclic
Given: A solvable finite group of order where are distinct primes. Every Sylow subgroup of is cyclic.
To prove: is metacyclic, with a cyclic normal subgroup and a cyclic quotient acting faithfully on it by conjugation.
Proof: Let be the Fitting subgroup of .
- : This follows from fact (2).
- is cyclic: First note that is nilpotent, so it is a direct product of its Sylow subgroups. Any Sylow subgroup of is a -subgroup of , hence is contained in a -Sylow subgroup. Thus, it is cyclic. This yields that is cyclic.
- acts faithfully on by conjugation: Since is cyclic, it is in particular abelian, so we get a well-defined action of the quotient group on (fact (4)). The kernel of this is , but as we concluded in step (1), this is trivial. Thus, we have a faithful action of on .
- is cyclic: Since is cyclic, its automorphism group is abelian, and from the previous step, is isomorphic to a subgroup of . Thus, is abelian. On the other hand, all its Sylow subgroups are cyclic, since these are quotients of Sylow subgroups of , which are cyclic. Thus, is itself cyclic.
The conclusions of steps (2), (3), and (4) complete the proof.
- 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