Center is local powering-invariant: Difference between revisions
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property = local powering-invariant subgroup}} | property = local powering-invariant subgroup}} | ||
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==Statement== | ==Statement== | ||
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* [[Fixed-point subgroup of a subgroup of the automorphism group implies local powering-invariant]] | * [[Fixed-point subgroup of a subgroup of the automorphism group implies local powering-invariant]] | ||
===Similar facts=== | |||
* [[Upper central series members are local powering-invariant in nilpotent group]] | |||
===Analogues in other algebraic structures=== | |||
* [[Center is local powering-invariant in Lie ring]] | |||
===Opposite facts=== | |||
* [[Center not is quotient-local powering-invariant]] | |||
* [[Derived subgroup not is local powering-invariant]] | |||
* [[Second center not is local powering-invariant in solvable group]] | |||
* [[Characteristic not implies powering-invariant]] | |||
==Facts used== | ==Facts used== | ||
Latest revision as of 18:09, 30 July 2013
This article gives the statement, and possibly proof, of the fact that for any group, the subgroup obtained by applying a given subgroup-defining function (i.e., center) always satisfies a particular subgroup property (i.e., local powering-invariant subgroup)}
View subgroup property satisfactions for subgroup-defining functions View subgroup property dissatisfactions for subgroup-defining functions
Statement
The center of a group is a local powering-invariant subgroup. Explicitly, suppose is a group and is the center. Suppose and is a natural number such that there is a unique satisfying . Then, .
Related facts
Generalizations
Similar facts
Analogues in other algebraic structures
Opposite facts
- Center not is quotient-local powering-invariant
- Derived subgroup not is local powering-invariant
- Second center not is local powering-invariant in solvable group
- Characteristic not implies powering-invariant
Facts used
Proof
Given: Group with center . Element and natural number such that there exists a unique satisfying .
To prove: . In other words, for all .
Proof: We have by Fact (1) that:
Simplifying further, we get that:
where we use that . Since is the unique element of whose <mah>n^{th}</math> power is , the above forces that .