An algebraic system $A$ is simple if the only congruences on it are $A\times A$ and $\Delta$ 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 $N_i\cap N_j=\varnothing$ 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, $[a^{-1}]=[1][a^{-1}]=[a][a^{-1}]=[aa^{-1}]=[1]$ so $a^{1}\in N$ $N$ is a subgroup of $G$ Furthermore, if $c\in G$ $[cac^{-1}]=[c][a][c^{-1}]= [c][1][c^{-1}]= [cc^{-1}]=[1]$ so that $cac^{-1}\in N$ $N$ 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 $\Delta$ which corresponds to $G$ having $G$ and $\langle 1\rangle$ 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 $\langle 1\rangle$ the only normal subgroups of $G$ Therefore, a simple group is a simple algebra.

Remark. Any simple algebraic system is subdirectly irreducible.