Powering-invariance is not quotient-transitive: Difference between revisions

Latest revision as of 16:00, 19 December 2014

This article gives the statement, and possibly proof, of a subgroup property (i.e., powering-invariant subgroup) not satisfying a subgroup metaproperty (i.e., quotient-transitive subgroup property).
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Statement

It is possible to have groups $H \leq K \leq G$ such that $H$ is a powering-invariant normal subgroup of $G$ and $K / H$ is a powering-invariant subgroup of the quotient group $G / H$ , but $K$ is not powering-invariant in $G$ .

Related facts

Proof

The proof idea is follows: use the construction in the reference for $H$ . Now take $K$ as a subgroup containing $H$ such that $K / H$ is a finite cyclic group of order $n > 1$ . Now:

$H$ is powering-invariant in $G$ by construction, since both $G$ and $H$ are rationally powered groups.
$K / H$ is powering-invariant in $G / H$ since $G / H$ is not powered over any prime.
$K$ is not powering-invariant in $G$ : For instance, an element of $K$ whose image in $K / H$ generates the latter group does not have a $n^{t h}$ root in $K$ .

References

MathOverflow question: Normal subgroup that is invariant under powering such that the quotient group is not invariant

@@ Line 14: / Line 14: @@
 ==Proof==
-The proof idea is follows: use the construction in the reference for <math>H</math>. Now take <math>K</math> as a subgroup containing <math>H</math> such that <math>K/H</math> is a finite cyclic group of order greater than 1. Now:
+The proof idea is follows: use the construction in the reference for <math>H</math>. Now take <math>K</math> as a subgroup containing <math>H</math> such that <math>K/H</math> is a finite cyclic group of order <math>n > 1</math>. Now:
 * <math>H</math> is powering-invariant in <math>G</math> by construction, since both <math>G</math> and <math>H</math> are [[rationally powered group]]s.
 * <math>K/H</math> is powering-invariant in <math>G/H</math> since <math>G/H</math> is not powered over ''any'' prime.
-* <math>K</math> is not powering-invariant in <math>G</math>: For instance, any element in the coset of <math>H</math> in <math>K</math> other than <math>H</math> cannot have a square root in <math>K</math>, even though <math>G</math> is 2-powered.
+* <math>K</math> is not powering-invariant in <math>G</math>: For instance, an element of <math>K</math> whose image in <math>K/H</math> generates the latter group does not have a <math>n^{th}</math> root in <math>K</math>.
 ==References==
 * {{mathoverflow|number = 121552|title = Normal subgroup that is invariant under powering such that the quotient group is not invariant}}