# Square map is endomorphism not implies abelian for loop

From Groupprops

## Statement

It is possible to have a loop in which the square map (where is the product of with itself) is an endomorphism but where is not a commutative loop, i.e., there exist such that .

## Related facts

### Opposite facts

- Square map is endomorphism iff abelian (for group)
- Square map is endomorphism iff abelian for diassociative loop

### Similar facts

## Proof

`Further information: loop of order five and exponent two`

We can construct an example non-abelian loop that has order five and where every element squares to the identity. Clearly, here, the square map is an endomorphism. The loop has the following multiplication table:

1 | 2 | 3 | 4 | 5 | |
---|---|---|---|---|---|

1 | 1 | 2 | 3 | 4 | 5 |

2 | 2 | 1 | 4 | 5 | 3 |

3 | 3 | 5 | 1 | 2 | 4 |

4 | 4 | 3 | 5 | 1 | 2 |

5 | 5 | 4 | 2 | 3 | 1 |