Moufang loop

Remarks

In a loop, this identity is equivalent to each of $x(z(yz)) = ((xz)y)z$, $(zx)(yz) = (z(xy))z$ and $(zx)(yz) = z((xy)z)$. A loop is Moufang exactly when it is both a left and a right Bol loop.

Example: the nonzero octonions $\mathbb{O} \setminus \set{0}$ under multiplication. They form a loop with identity $1$, since $\mathbb{O}$ is a division algebra. The identity holds because the octonions satisfy the Moufang identities as an alternative algebra. It is not a group, because octonion multiplication is not associative.

Non-example: the loop $L = \set{e, a, b, c, d}$ with the multiplication table below, where the row gives the left factor.

Each element occurs once in each row and column, so $L$ is a loop with identity $e$. Take $z = a$, $x = e$, $y = b$: then $z(x(zy)) = a(ab) = ac = d$, while $((zx)z)y = (aa)b = eb = b$.