Difference between revisions of "Commuting fraction more than five-eighths implies abelian"
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It is likely that this statement, being quite simple, was known to many people since the beginnings of group theory and/or rediscovered a number of times. The first written proof
It is likely that this statement, being quite simple, was known to many people since the beginnings of group theory and/or rediscovered a number of times. The first written proof in a paper by Erdos and Turan.
Latest revision as of 04:46, 5 January 2018
This article is about a result whose hypothesis or conclusion has to do with the fraction of group elements or tuples of group elements satisfying a particular condition.The fraction involved in this case is 5/8
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It is likely that this statement, being quite simple, was known to many people since the beginnings of group theory and/or rediscovered a number of times. The first written proof appears in a paper by Erdos and Turan.
For a finite group in terms of commuting fractions
In particular, if we define:
Note that since the commuting fraction of a finite abelian group is , this means that the commuting fraction of a finite group cannot take any value strictly between and .
For a finite group in terms of conjugacy classes
For a FZ-group
The fact has applications to the abelianness testing problem, and is the basis for the randomized black-box group algorithm for abelianness testing.
- Commuting fraction equal to five-eighths iff inner automorphism group is Klein four-group
- Commuting fraction more than half implies nilpotent
- Fixed-class tuple fraction is bounded away from one for groups not of that class plays a similar role for the fixed-class nilpotency testing problem
- Randomized black-box group algorithm for normality testing has false positive rate of at most three-fourths plays a similar role for the normality testing problem
- Cyclic over central implies abelian
- Lagrange's theorem in the special form subgroup of size more than half is whole group
Proof for a finite non-abelian group
Given: A finite non-abelian group .
To prove: The commuting fraction of is at most .
Proof: Let be the center of .
|Step no.||Assertion||Given data used||Facts used||Previous steps used||Explanation|
|1||is not cyclic||is non-abelian||Fact (1)||--||By fact (1), if were cyclic, would be abelian, a contradiction.|
|2||is at least||--||Step (1)||Any group of order 1, 2, or 3 is cyclic, forcing to have size at least .|
|3||For all elements in and not in , the quotient is at most||Fact (2)||--||If is not in , then is a proper subgroup of . By fact (2), we get that .|
|4||The commuting fraction of is||--||--||By definition. Explicitly, the commuting fraction is . Those that are in can be separated out and we get the summation in the form presented.|
|5||The commuting fraction of is at most||--||Steps (3) and (4)||By Step (3), for each . The sum is thus bounded by the size of times this individual bound.|
|6||The commuting fraction of is at most||--||Step (5)||Algebraic simplification|
|7||The commuting fraction of is at most||Fact (2)||Steps (2), (6)||By Fact (2), which by Step (2) is at most 1/4. Plugging this into Step (6), we get that the commuting fraction of is at most .|