# Square map is endomorphism not implies abelian for loop

## Statement

It is possible to have a loop $(L,*)$ in which the square map $x \mapsto x^2$ (where $x^2$ is the product of $x$ with itself) is an endomorphism but where $L$ is not a commutative loop, i.e., there exist $x,y \in L$ such that $x * y = y * x$.

## Proof

Further information: loop of order five and exponent two

We can construct an example non-abelian loop $L$ 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