In this question, all sets are finite. A semigroup is a set equipped with an associative binary operation, which I will call $\circ$. The tranformation semigroup on a set $X$ is the semigroup of maps $X\to X$ with function composition as the semigroup operation. The degree of a semigroup $S$ is the smallest integer $n$ such that $S$ is a subsemigroup (a subset closed under the semigroup operation) of the transformation semigroup on a set of order $n$.

The question is whether there exists $n$ such that for all semigroups of order at least $n$, the degree of $S$ is at most $|S|$.

Every semigroup $S$ can be extended to a semigroup with identity (also called a monoid) $M_S=S\cup\{1\}$, where the semigroup operation $\circ$ is extended by setting $1\circ s=s\circ 1=s$ for all $s\in S$. A semigroup $M$ with identity has degree at most $|M|$, because it is isomorphic to its "regular representation", the semigroup of functions $f_m\colon M\to M$ where $f_m(m')=m\circ m'$. See http://en.wikipedia.org/wiki/Transformation_semigroup#Cayley_representation

Restricting to $S$, this shows that the degree of $S$ is at most $|S|+1$. This cannot be improved to $|S|$. Let $S=\{a,b\}$ and $s\circ s'=a$ for all $s,s'\in S$. Suppose that $S$ was a subsemigroup of the four functions $\{0,1\}\to \{0,1\}$ under composition. Since $S$ does not have an identity elements, we can rule out using the identity function or the swapping function $t\mapsto 1-t$. That leaves only the constant maps. But the constant maps give a different semigroup operation, $s\circ s'=s$.

Clearly a counterexample cannot have an identity element. It is known that there are groups $G$ (and therefore semigroups) of degree $|G|$, for example cyclic groups of order $p^n$ where $p$ prime. (More is known.)

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    $\begingroup$ Have you also tried $n=3$? $\endgroup$
    – Berci
    Oct 19, 2012 at 18:23
  • $\begingroup$ It seems that the "$s\circ s'=a$" counterexample generalises to $S=\{a,b,c\}$, by checking cases. A function $f\colon\{0,1,2\}\to\{0,1,2\}$ representing $b$ must satisfy $f\neq f^2=f^3$, and the only options are $0\mapsto 1\mapsto 2\mapsto 2$ and conjugate functions. Then there are no functions left that can represent $c$. $\endgroup$ Oct 19, 2012 at 19:36

1 Answer 1


The answer is negative. Let $\mathcal{T}_n$ be the transformation semigroup on $n$ elements and let $S_n = \{s, s^2, \ldots, s^n\}$ be the semigroup defined by $s^{n+1} = s^n$. I will use the following consequence of Green's lemma:

Lemma 1. Let $S$ be a finite semigroup and let $s$ and $t$ be two $\mathcal{J}$-equivalent elements of $S$. If $ts = t$, then $s$ is idempotent.

Proposition. Let $n > 1$. Then the semigroup $S_n$ is not a subsemigroup of $\mathcal{T}_n$.

Proof. Suppose by contradiction there is an element $s \in \mathcal{T}_n$ such that $s^{n+1} = s^n$ but $s^i \not= s^j$ for $1 \leqslant i < j \leqslant n$. I claim that $s^n <_\mathcal{J} s^{n-1} <_\mathcal{J} \cdots <_\mathcal{J} s$. It is clear that $s^{i+1} \leqslant_\mathcal{J} s^i$ for $1 \leqslant i \leqslant n-1$. Moreover, if $s^{i+1} \mathrel{\mathcal{J}} s^i$, then by Lemma 1 applied to $t = s^i$, $s$ is idempotent, a contradiction. However, $\mathcal{T}_n$ has exactly $n$ $\mathcal{J}$-classes: $J_1 <_\mathcal{J} J_2 <_\mathcal{J} \cdots <_\mathcal{J} J_n$, where each $J_k$ is the set of elements of rank $k$. Thus $s$ necessarily belongs to $J_n$, the group of units of $\mathcal{T}_n$. It follows that $s^n = 1$ and $s^{n+1} = s$, a contradiction.

  • $\begingroup$ $s$ being idempotent isn't a contradiction. In fact, $\mathcal{S}_1$ is a subsemigroup of $\mathcal{T}_1$. $\endgroup$
    – Jakobian
    Dec 23, 2020 at 22:59
  • $\begingroup$ @Jakobian Thanks. I have added the condition $n > 1$. $\endgroup$
    – J.-E. Pin
    Dec 24, 2020 at 4:30

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