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Homesimple algebraic system

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# simple algebraic system

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 $i\neq j$. 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.

## Mathematics Subject Classification

08A30*no label found*

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