Powering-invariant not implies divisibility-closed: Difference between revisions

Latest revision as of 17:15, 15 February 2013

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

It is possible to have a group $G$ and a subgroup $H$ such that:

$H$ is a powering-invariant subgroup of $G$ : If $n$ is a natural number such that every element of $G$ has a unique $n^{t h}$ root, then every element of $H$ has a unique $n^{t h}$ root within $H$ .
$H$ is not a divisibility-closed subgroup of $G$ : There exists a natural number $n$ such that every element of $G$ has a $n^{t h}$ root (not necessarily unique) but not every element of $H$ has a $n^{t h}$ root within $H$ .

Related facts

Center not is divisibility-closed

Proof

Proof idea

The key fact is that any finite subgroup of a group must be powering-invariant, but it need not be divisibility-closed. We will construct an example where the subgroup is finite.

Proof details

For any prime number $p$ :

Let $G$ be the $p$ -quasicyclic group.
Let $H$ be the subgroup comprising the elements of order 1 or $p$ .

Clearly:

$H$ , being finite, is powering-invariant (in fact, both $G$ and $H$ are powered over precisely the set of primes other than $p$ ).
However, $H$ is not divisibility-closed: $G$ is $p$ -divisible, but $H$ is not.

@@ Line 1: / Line 1: @@
 {{subgroup property non-implication|
 stronger = powering-invariant subgroup|
-weaker = divisibility-invariant subgroup}}
+weaker = divisibility-closed subgroup}}
 ==Statement==
-It is possible to have a [[group]] <math>G</math> and a [[subgroup]] <math>H</math>
+It is possible to have a [[group]] <math>G</math> and a [[subgroup]] <math>H</math> such that:
+# <math>H</math> is a [[powering-invariant subgroup]] of <math>G</math>: If <math>n</math> is a natural number such that every element of <math>G</math> has a unique <math>n^{th}</math> root, then every element of <math>H</math> has a unique <math>n^{th}</math> root within <math>H</math>.
+# <math>H</math> is ''not'' a [[divisibility-closed subgroup]] of <math>G</math>: There exists a natural number <math>n</math> such that every element of <math>G</math> has a <math>n^{th}</math> root (not necessarily unique) but not every element of <math>H</math> has a <math>n^{th}</math> root within <math>H</math>.
+==Related facts==
+* [[Center not is divisibility-closed]]
 ==Proof==
@@ Line 11: / Line 18: @@
 ===Proof idea===
-The key fact is that any finite subgroup of a group must be powering-invariant, but it need not be divisibility-invariant. We will construct an example where the subgroup is finite.
+The key fact is that any finite subgroup of a group must be powering-invariant, but it need not be divisibility-closed. We will construct an example where the subgroup is finite.
 ===Proof details===
-For any prime number <math>p</math>, let <math>G</math> be the <math>p</math>-[[quasicyclic group]]. Let <math>H</math> be the subgroup comprising the elements of order 1 or <math>p</math>. Clearly, <math>H</math>, being finite, is powering-invariant (in fact, both <math>G</math> and <math>H</math> are powered over precisely the set of primes other than <math>p</math>). However, <math>H</math> is not divisibility-invariant: <math>G</math> is <math>p</math>-divisible, but <math>H</math> is not.
+For any prime number <math>p</math>:
+* Let <math>G</math> be the <math>p</math>-[[quasicyclic group]].
+* Let <math>H</math> be the subgroup comprising the elements of order 1 or <math>p</math>.
+Clearly:
+* <math>H</math>, being finite, is powering-invariant (in fact, both <math>G</math> and <math>H</math> are powered over precisely the set of primes other than <math>p</math>).
+* However, <math>H</math> is not divisibility-closed: <math>G</math> is <math>p</math>-divisible, but <math>H</math> is not.