# Difference between revisions of "Commuting fraction more than five-eighths implies abelian"

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==History== | ==History== | ||

− | 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 appears in a paper by Erdos and Turan. |

==Statement== | ==Statement== |

## 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

View other such statements

## Contents

## History

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.

## Statement

### For a finite group in terms of commuting fractions

Suppose is a finite group such that the commuting fraction of is more than . Then, is an abelian group, and in particular, a finite abelian group.

In particular, if we define:

Then:

abelian

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

Suppose is a finite group and is the number of conjugacy classes in . Then:

abelian.

### For a FZ-group

Suppose is a FZ-group whose commuting fraction is more than . Then, is an abelian group.

## Related facts

### Applications

The fact has applications to the abelianness testing problem, and is the basis for the randomized black-box group algorithm for abelianness testing.

### Similar facts

- 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

## Facts used

- Cyclic over central implies abelian
- Lagrange's theorem in the special form subgroup of size more than half is whole group

## Proof

### 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 . |