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An algebraic system  is simple if the only congruences on it are  and  , the diagonal relation.
For example, let's find out what are the simple algebras in the class of groups. Let  be a group that is simple in the sense defined above.
First, what are the congruences on  ? A congruence  on  is a subgroup of  and an equivalence relation on  at the same time. As an equivalence relation,  corresponds to a partition of  in the following manner:
 and
 , where
 for  . Each of the  is an equivalence class of  . Let  be the equivalence class containing  . If  , then
![$ [a]=[b]=[1]$](http://images.planetmath.org:8080/cache/objects/9013/l2h/img21.png "$ [a]=[b]=[1]$") , so that
![$ [ab]=[a][b]=[1][1]=[1]$](http://images.planetmath.org:8080/cache/objects/9013/l2h/img22.png "$ [ab]=[a][b]=[1][1]=[1]$") , or  . In addition,
![$ [a^{-1}]=[1][a^{-1}]=[a][a^{-1}]=[aa^{-1}]=[1]$](http://images.planetmath.org:8080/cache/objects/9013/l2h/img24.png "$ [a^{-1}]=[1][a^{-1}]=[a][a^{-1}]=[aa^{-1}]=[1]$") , so
 .  is a subgroup of  . Furthermore, if  ,
![$ [cac^{-1}]=[c][a][c^{-1}]= [c][1][c^{-1}]= [cc^{-1}]=[1]$](http://images.planetmath.org:8080/cache/objects/9013/l2h/img29.png "$ [cac^{-1}]=[c][a][c^{-1}]= [c][1][c^{-1}]= [cc^{-1}]=[1]$") , so that
 ,  is a normal subgroup of  . Conversely, given a normal subgroup  of  , forming left (right) cosets  of  , and taking
 gives us the congruence  on  .
Now, if  is simple, then this says that the only congruences on  are  and  , which corresponds to  having  and
 as the only normal subgroups. So,  as a simple algebra is just a simple group. Conversely, if  is a simple group, then the only congruences on  are those corresponding to  and
 , the only normal subgroups of  . Therefore, a simple group is a simple algebra.
Remark. Any simple algebraic system is subdirectly irreducible.
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