Factor congruence

Remarks

For such a pair, the map $a \mapsto (a/\theta, a/\theta^*)$ is an isomorphism from $\mathbf{A}$ onto the direct product $\mathbf{A}/\theta \times \mathbf{A}/\theta^*$. Conversely, if $\mathbf{A} = \mathbf{B} \times \mathbf{C}$, the kernels of the two projections are complementary factor congruences. The conditions say that $\theta$ and $\theta^*$ are complements in the congruence lattice and that they permute.

Example: on the group $\Z/6\Z$, take $\theta$ to be congruence modulo $2$ and $\theta^*$ congruence modulo $3$. An element congruent to $0$ modulo $2$ and modulo $3$ is $0$, so clause 1 holds. For any $a, c$ there is $b$ with $b \equiv a \pmod 2$ and $b \equiv c \pmod 3$, by the Chinese remainder theorem, so clause 2 holds.

Non-example: on $\Z/4\Z$, congruence modulo $2$ is not a factor congruence. The congruences of $\Z/4\Z$ are $\Delta$, congruence modulo $2$, and $\nabla$. Only $\theta^* = \Delta$ satisfies clause 1, and then $\theta \circ \theta^* = \theta \ne \nabla$.

The congruences $\Delta_A$ and $\nabla_A$ are always factor congruences, complementary to each other.