Cyclic automorphism group not implies cyclic: Difference between revisions

Latest revision as of 16:53, 25 June 2013

This article gives the statement and possibly, proof, of a non-implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., group whose automorphism group is cyclic) need not satisfy the second group property (i.e., cyclic group)
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Statement

It is possible to have a group that is an [group whose automorphism group is cyclic]] (i.e., the automorphism group is a cyclic group) but the group itself is not a cyclic group.

Related facts

Proof

Further information: group of rational numbers with square-free denominators

Let $G$ be the group of rational numbers with square-free denominators, i.e., $G$ is the subgroup of the additive group of rational numbers comprising those rational numbers that, when written in reduced form, have denominators that are square-free numbers, i.e., there is no prime number $p$ for which $p^{2}$ divides the denominator. Then:

The only non-identity automorphism of $G$ is the negation map, so the automorphism group is cyclic group:Z2, and is hence cyclic.
$G$ is not a cyclic group. In fact, it is not even a finitely generated group because any finite subset of $G$ can only cover finitely many primes in their denominators. It is, however, a locally cyclic group: any finitely generated subgroup is cyclic.

@@ Line 1: / Line 1: @@
 {{group property non-implication|
-stronger = aut-cyclic group|
+stronger = group whose automorphism group is cyclic|
 weaker = cyclic group}}
 ==Statement==
-It is possible to have a [[group]] that is an [[aut-cyclic group]] (i.e., the [[automorphism group]] is a [[cyclic group]]) but the group itself is not a [[cyclic group]].
+It is possible to have a [[group]] that is an [group whose automorphism group is cyclic]] (i.e., the [[automorphism group]] is a [[cyclic group]]) but the group itself is not a [[cyclic group]].
 ==Related facts==
-* [[Finite and aut-cyclic implies cyclic]]
+* [[Finite and cyclic automorphism group implies cyclic]]
-* [[Aut-cyclic implies abelian]]
+* [[Cyclic automorphism group implies abelian]]
-* [[Finite abelian and aut-abelian implies cyclic]]
+* [[Finite abelian and abelian automorphism group implies cyclic]]
-* [[Abelian and aut-abelian not implies locally cyclic]]
+* [[Abelian and abelian automorphism group not implies locally cyclic]]
 ==Proof==
-Let <math>G</math> be the subgroup of the additive [[group of rational numbers]] comprising those rational numbers that, when written in reduced form, have denominators that are [[square-free number]]s, i.e., there is no [[prime number]] <math>p</math> for which <math>p^2</math> divides the denominator. Then:
+{{further|[[particular example::group of rational numbers with square-free denominators]]}}
+Let <math>G</math> be the [[group of rational numbers with square-free denominators]], i.e., <math>G</math> is the subgroup of the additive [[group of rational numbers]] comprising those rational numbers that, when written in reduced form, have denominators that are [[square-free number]]s, i.e., there is no [[prime number]] <math>p</math> for which <math>p^2</math> divides the denominator. Then:
 * The only non-identity automorphism of <math>G</math> is the negation map, so the automorphism group is [[cyclic group:Z2]], and is hence cyclic.
 * <math>G</math> is not a cyclic group. In fact, it is not even a [[finitely generated group]] because any finite subset of <math>G</math> can only cover finitely many primes in their denominators. It is, however, a [[locally cyclic group]]: any finitely generated subgroup is cyclic.