simple algebraic system

An algebraic system $A$ is simple if the only congruences on it are $A\times A$ and $\mathrm{\Delta }$, the diagonal relation.

For example, let’s find out what are the simple algebras in the class of groups. Let $G$ be a group that is simple in the sense defined above.

First, what are the congruences on $G$? A congruence $C$ on $G$ is a subgroup of $G\times G$ and an equivalence relation on $G$ at the same time. As an equivalence relation, $C$ corresponds to a partition of $G$ in the following manner: $G={\bigcup }_{i\in I}{N}_i$ and $C={\bigcup }_{i\in I}{N}_i^2$, where $N_i\cap N_j=\mathrm{\varnothing }$ for $i\ne j$. Each of the $N_i$ is an equivalence class of $C$. Let $N$ be the equivalence class containing $1$. If $a,b\in N$, then $[a]=[b]=[1]$, so that $[ab]=[a][b]=[1][1]=[1]$, or $ab\in N$. In addition, $[a^{-1}]=[1][a^{-1}]=[a][a^{-1}]=[a{a}^{-1}]=[1]$, so $a^1\in N$. $N$ is a subgroup of $G$. Furthermore, if $c\in G$, $[ca{c}^{-1}]=[c][a][c^{-1}]=[c][1][c^{-1}]=[c{c}^{-1}]=[1]$, so that $ca{c}^{-1}\in N$, $N$ is a normal subgroup of $G$. Conversely, given a normal subgroup $N$ of $G$, forming left (right) cosets $N_i$ of $N$, and taking $C={\bigcup }_{i\in I}{N}_i^2$ gives us the congruence $C$ on $G$.

Now, if $G$ is simple, then this says that the only congruences on $G$ are $G\times G$ and $\mathrm{\Delta }$, which corresponds to $G$ having $G$ and $⟨1⟩$ as the only normal subgroups. So, $G$ as a simple algebra is just a simple group. Conversely, if $G$ is a simple group, then the only congruences on $G$ are those corresponding to $G$ and $⟨1⟩$, the only normal subgroups of $G$. Therefore, a simple group is a simple algebra.

Remark. Any simple algebraic system is subdirectly irreducible.