Directly indecomposable algebra

Remarks

Equivalently, the only factor congruences of $\mathbf{A}$ are $\Delta_A$ and $\nabla_A$. Every finite algebra is isomorphic to a direct product of finitely many directly indecomposable algebras, by induction on the number of elements.

Example: an algebra $\mathbf{A}$ with a prime number $p$ of elements is directly indecomposable. If $\mathbf{A} \cong \mathbf{B} \times \mathbf{C}$, then $p = \abs{B} \cdot \abs{C}$, so $\abs{B} = 1$ or $\abs{C} = 1$.

Non-example: the group $\Z/6\Z$ is not directly indecomposable. The map $k \mapsto (k \bmod 2, k \bmod 3)$ is an isomorphism onto $\Z/2\Z \times \Z/3\Z$, and both factors have at least two elements.

Every subdirectly irreducible algebra is directly indecomposable, but not conversely: the ring $\Z$ is directly indecomposable and not subdirectly irreducible.