Frobenius conjecture on nth roots: Difference between revisions
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{{conjecture | {{conjecture|finite groups}} | ||
==Statement== | ==Statement== | ||
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==Related facts== | ==Related facts== | ||
* [[Exactly n elements of order dividing n in a finite solvable group implies the elements form a subgroup]]: This is precisely the Frobenius conjecture in the case of [[finite solvable group]]s. | |||
* [[Number of nth roots is a multiple of n]] | * [[Number of nth roots is a multiple of n]] | ||
* [[Number of nth roots of any conjugacy class is a multiple of n]] | * [[Number of nth roots of any conjugacy class is a multiple of n]] | ||
* [[At most n elements of order dividing n implies every finite subgroup is cyclic]] | * [[At most n elements of order dividing n implies every finite subgroup is cyclic]] |
Latest revision as of 22:49, 11 November 2015
This article is about a conjecture in the following area in/related to group theory: finite groups. View all conjectures and open problems
Statement
Suppose is a finite group and is a natural number dividing the order of . Suppose the number of roots in , i.e., the number of elements such that , is exactly .
The Frobenius conjecture on nth roots states that in that case, that set of roots must be a subgroup of .
Related facts
- Exactly n elements of order dividing n in a finite solvable group implies the elements form a subgroup: This is precisely the Frobenius conjecture in the case of finite solvable groups.
- Number of nth roots is a multiple of n
- Number of nth roots of any conjugacy class is a multiple of n
- At most n elements of order dividing n implies every finite subgroup is cyclic