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 </math>x^2</math> 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