# Square map is endomorphism iff abelian

This article describes an easy-to-prove fact about basic notions in group theory, that is not very well-known or important in itself

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## Contents

## Statement

Let be a group and be the square map of defined as . Then, is an endomorphism of (i.e., ) if and only if is abelian.

Another way of putting it is that is 2-abelian if and only if it is abelian.

## Related facts

### Applications

- Exponent two implies abelian: If the exponent of a group is 2 (i.e., the group is nontrivial and every non-identity element has order two) then the group is abelian. The analogous statement is not true for any other prime number, i.e., there can be a non-abelian group of prime exponent. The standard example for an odd prime is prime-cube order group:U(3,p) of order .

### Majority criterion

### Other power maps

The power map for a fixed integer is termed a universal power map, and if it is also an endomorphism, it is termed a universal power endomorphism and the group is termed a n-abelian group. This statement gives a necessary and sufficient condition for a group where gives an endomorphism. Here are results for other values of .

- n-abelian iff (1-n)-abelian
- The set of for which is -abelian is termed the exponent semigroup of . It is a submonoid of the multiplicative monoid of integers.
- abelian implies n-abelian for all n
- n-abelian implies every nth power and (n-1)th power commute
- n-abelian implies n(n-1)-central
- n-abelian iff abelian (if order is relatively prime to n(n-1))
- nth power map is surjective endomorphism implies (n-1)th power map is endomorphism taking values in the center
- (n-1)th power map is endomorphism taking values in the center implies nth power map is endomorphism
- Frattini-in-center odd-order p-group implies p-power map is endomorphism
- Frattini-in-center odd-order p-group implies (mp plus 1)-power map is automorphism
- Characterization of exponent semigroup of a finite p-group
- Alperin's structure theorem for n-abelian groups

Value of (note that the condition for is the same as the condition for ) | Characterization of -abelian groups | Proof | Other related facts |
---|---|---|---|

0 | all groups | obvious | |

1 | all groups | obvious | |

2 | abelian groups only | 2-abelian iff abelian | endomorphism sends more than three-fourths of elements to squares implies abelian |

-1 | abelian groups only | -1-abelian iff abelian | |

3 | 3-abelian group means: 2-Engel group and derived subgroup has exponent dividing three | Levi's characterization of 3-abelian groups | cube map is surjective endomorphism implies abelian, cube map is endomorphism iff abelian (if order is not a multiple of 3), cube map is endomorphism implies class three |

-2 | same as for 3-abelian | (based on n-abelian iff (1-n)-abelian) |

### Related facts for Lie rings

Here are some related facts for Lie rings:

- Multiplication by n map is a derivation iff derived subring has exponent dividing n
- Multiplication by n map is an endomorphism iff derived subring has exponent dividing n(n-1)

### Opposite facts for other algebraic structures

Statement | Algebraic structure | What step of the proof fails? | Comment |
---|---|---|---|

Square map is endomorphism not implies abelian for loop | loop | The reparenthesization in Step (3) of the proof below, that requires associativity. | In fact, it is possible to have a noncommutative loop of exponent two. |

Square map is endomorphism not implies abelian for monoid | monoid | The cancellation in Step (4), which requires that we are working over a cancellative monoid. |

## Facts used

- Associative implies generalized associative: Basically this says that in a group, we can drop and rearrange parentheses at will.
- Invertible implies cancellative in monoid. Since every element of a group is invertible, cancellation is valid in groups.
- Abelian implies universal power map is endomorphism

## Proof

### From square map being endomorphism to abelian

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

**Given**: A group such that the map is an endomorphism, i.e., for all .

**To prove**: for all .

**Proof**: We let be arbitrary elements of .

Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation | What algebraic assumptions does this use? |
---|---|---|---|---|---|---|

1 | -- | square map is endomorphism | -- | -- | None, works over any magma | |

2 | -- | Step (1) | -- | None, just using definition of square. Works over any magma. | ||

3 | Fact (1) | Step (2) | Reparenthesize | The reparenthesization requires associativity of expressions involving two variables. It works over any semigroup or monoid and even more generally over any diassociative magma. | ||

4 | Fact (2) | Step (3) | Cancel the right-most from both sides, then the left-most from both sides. | The cancellation requires that we are working in a cancellative magma, such as a cancellative monoid or a quasigroup or loop. |

### From abelian to square map being endomorphism

This follows directly from fact (3).