Mal'cev's theorem on congruence permutability

Remarks

Suppose $p$ is a Mal'cev term, $A \in \mathcal{V}$, and $a \mathrel{\alpha} b \mathrel{\beta} c$ for congruences $\alpha$ and $\beta$ of $A$. Put $d = p(a, b, c)$. Then $d \mathrel{\beta} p(a, b, b) = a$ and $d \mathrel{\alpha} p(b, b, c) = c$, so $a \mathrel{\beta} d \mathrel{\alpha} c$, and the two compositions are equal. Conversely, work in the free algebra $F$ of $\mathcal{V}$ on $x, y, z$, with $\alpha$ generated by $(x, y)$ and $\beta$ generated by $(y, z)$. Since $x \mathrel{\alpha} y \mathrel{\beta} z$, permutability gives $w = p(x, y, z) \in F$ with $x \mathrel{\beta} w \mathrel{\alpha} z$, and identifying $z$ with $y$, or $y$ with $x$, turns these relations into the two identities.

Example: groups have the Mal'cev term $x y^{-1} z$, so the congruences of every group permute; this is the familiar fact that $MN = NM$ for normal subgroups. Lattices have no Mal'cev term, since the three-element chain has two congruences that do not permute.

Going deeper

For a student

This is the first Mal'cev condition, and it is the model for later ones, such as Jónsson's characterisation of congruence distributivity and Day's characterisation of congruence modularity. The proof pattern is general: an implication in the free algebra on few generators gives terms, and terms give the property in every algebra. Quasigroups, with $p(x, y, z) = (x / (y \backslash y)) \cdot (y \backslash z)$, also have a Mal'cev term, so their congruences permute.