Cyclic normal implies potentially verbal in finite: Difference between revisions
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Suppose <math>K</math> is a [[finite group]] and <math>H</math> is a [[cyclic normal subgroup]] of <math>K</math>. In other words, <math>H</math> is a [[normal subgroup]] of <math>K</math> that is also a [[cyclic group]]. Then, <math>H</math> is a [[potentially verbal subgroup]] of <math>G</math>: there exists a group <math>G</math> containing <math>K</math> such that <math>H</math> is a [[verbal subgroup]] of <math>G</math>. | Suppose <math>K</math> is a [[finite group]] and <math>H</math> is a [[cyclic normal subgroup]] of <math>K</math>. In other words, <math>H</math> is a [[normal subgroup]] of <math>K</math> that is also a [[cyclic group]]. Then, <math>H</math> is a [[potentially verbal subgroup]] of <math>G</math>: there exists a group <math>G</math> containing <math>K</math> such that <math>H</math> is a [[verbal subgroup]] of <math>G</math>. In fact, we can choose <math>G</math> to itself be a [[finite group]]. | ||
==Related facts== | ==Related facts== | ||
Revision as of 21:09, 19 October 2009
This article gives the statement and possibly, proof, of an implication relation between two subgroup properties, when the big group is a finite group. That is, it states that in a Finite group (?), every subgroup satisfying the first subgroup property (i.e., Cyclic normal subgroup (?)) must also satisfy the second subgroup property (i.e., Potentially verbal subgroup (?)). In other words, every cyclic normal subgroup of finite group is a potentially verbal subgroup of finite group.
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Statement
Statement with symbols
Suppose is a finite group and is a cyclic normal subgroup of . In other words, is a normal subgroup of that is also a cyclic group. Then, is a potentially verbal subgroup of : there exists a group containing such that is a verbal subgroup of . In fact, we can choose to itself be a finite group.
