Square map is endomorphism not implies abelian for loop

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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.

Related facts

Opposite facts

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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