I am reading the book " Fundamentals of semigroup theory" by John M. Howie $\textbf{(Section 3.7)}$.

I want to prove $\textbf{Theorem 3.7.2}$

If $S$ is a finite congruence free semigroup without zero and if $|S| > 2$, then $S$ is a simple group.

Let $S$ be a finite congruence free semigroup without zero. Then $S = M[G; I, \Lambda, P]$, where $P$ is a regular matrix of order $\Lambda \times I$. We know that for every proper congruences on $S$, there exist a linked triple and for every linked triple there is a proper congruence.

A triplet $(N, K, T)$, where $N$ is a normal subgroup of $G$, $K$ is an equivalence relation on $I$ such that $K \subseteq \varepsilon_I$ and $T$ is an equivalence relation on $\Lambda$ such that $T \subseteq \varepsilon _\Lambda$ and $q_{\lambda \mu ij} \in N$, whenever $(i,j) \in K$ or $(\lambda , \mu ) \in T$

Firstly we shall show that $G$ is a simple group. If $N$ is a proper normal subgroup of $G$, then $(N, 1_I, 1_{\lambda} )$ is a linked triple and gives a congruence which distinct from universal and identical congruences. Hence if $S$ is congruence free , then $G$ is either simple or $G = \{e\}$. If $G = \{e\}$, then $|S| = |I \times \Lambda| > 2$ and so either $|I| =| \Lambda | =2$ or atleast one of $I$, $\Lambda$ (say $I$) has more then two element. In the first case we find the linked triples $(\{e\}, 1_I, \Lambda \times \Lambda)$ and $(\{e\}, I \times I, 1_{\Lambda} )$ are the linked triples gives non trivial congruences, while in the second case there exist an equivalences $K$ on $I$ such that $1_I \subset K \subset I \times I$ which gives a linked triple $(\{e\}, K, 1_{\Lambda})$ and this gives rises a non-trivial congruences. Hence if $S$ is congruence free, then $G \neq \{e\}$

But in the starting of the section we realize that

Since $(G, 1_I, 1_{\lambda})$ is a linked triplet and gives a non-identical proper congruences if $G \neq \{e \}$ and we assume that $S$ is a finite congruence free semigroup, so we get a contradiction.

I am confused. Please tell me where am i wrong. Thank you


closed as off-topic by RP_, Franz Lemmermeyer, Stefan Kohl, Max Horn, Mark Sapir May 1 '17 at 0:43

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The way to think about this is since S has no zero its minimal ideal is non-trivial. If it is not the whole semigroup you can factor out the minimal ideal to get a non trivial congruence.

So $S$ must be its minimal ideal. An easy consequence of Rees's theorem is that Greens relations $\mathcal L$ and $\mathcal R$ are congruences on a finite simple semigroup. When you factor out $\mathcal L$ you get a right zero semigroup. Thus either $S$ is a right zero semigroup or has a single $\mathcal L$-class. Every equivalence relation on a right zero semigroup is a congruence. Since your semigroup has more than two elements it cannot be a right zero semigroup. So there is one $\mathcal L$-class.

Dually, there is one $\mathcal R$-class. Thus $S$ is a group and hence a simple group.

Now you should check if $S$ has a zero and more than two elements then it must be a Rees Matrix semigroup over the trivial group with sandwich matrix containing no equal rows or columns and no zero rows or columns.


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