## Facts about finite cyclic semigroups

Let $S$ be a finite cyclic semigroup. Show that $S$ has a unique maximal subgroup and compute it. Compute the kernel of $S$. Under what circumstances is $S$ a group? Deduce that every finite semigroup contains an idempotent.

Let $S$ be a finite cyclic subgroup with generator $s$. Now the map $\varphi : \mathbb{N}^+ \rightarrow S$ given by $k \mapsto s^k$ is a surjective semigroup homomorphism (where $\mathbb{N}^+$ is a semigroup under $+$). Since $\varphi$ cannot be injective, there exist pairs of distinct positive natural numbers $m+1 > n$ such that $s^{m+1} = s^n$. Let $(n,m+1)$ be such that $m+1$ is minimal among these pairs. (For such $m+1$, there is a unique $n < m+1$ with $s^n = s^{m+1}$, since otherwise $m+1$ is not minimal.)

We now prove a lemma.

Lemma 1: If $n,m \in \mathbb{N}^+$ with $n \leq m$, then there exists a unique $d \in \mathbb{N}^+$ such that $n \leq d(m+1-n) < m+1$. Proof: (Existence) Note that $m+1-n > 0$. By the Division Algorithm in $\mathbb{N}$, there exist natural numbers (that is, nonnegative integers) $q$ and $r$ such that $n = q(m+1-n) + r$ and $0 \leq r < m+1-n$. Letting $-t = r - (m+1-n)$ and $d = q+1$, we have $n = d(m+1-n) - t$, so that $0 \leq d(m+1-n) - n = t < m+1-n$, and thus $n \leq d(m+1-n) < m+1$. (Uniqueness) Suppose $n \leq d(m+1-n) < m$ and $n \leq e(m+1-n) \leq m$, we have $(d-e)(m+1-n) = 0$, so that $d-e = 0$, and so $d = e$. $\square$

This number $d(m+1-n)$ will turn out to be very special.

Lemma 2: For all $i \in \mathbb{N}$ and $k \in \mathbb{N}^+$, we have $s^{k(m+1-n)}s^{n+i} = s^{n+i}$. Proof: We proceed by induction on $k$. For the base case $k = 1$, if $i = 0$, then $s^{m+1-n}s^{n+i} = s^{m+1-n+n}$ $= s^{m+1} = s^n$ $= s^{n+i}$. If $i \neq 0$, then $s^{m+1-n}s^{n+i} = s^{m+1+i}$ $= s^{m+1}s^i$ $= s^ns^i$ $= s^{n+i}$. Now for the inductive step, suppose the result holds for some $k$. Then $s^{(k+1)(m+1-n)}s^{n+i} = s^{m+1-n}s^{k(m+1-n)}s^{n+1}$ $= s^{m+1-n}s^{n+i}$ $= s^{n+i}$. So the result holds for all $k \geq 1$. $\square$

Corollary 2.1: If $a,b \geq n$ and $a \equiv b$ mod $m+1-n$, then $s^a = s^b$. Proof: Without loss of generality, say $a > b$. Now $a-b = q(m+1-n)$ for some $q \geq 1$, and thus $a = q(m+1-n) + b$ and $b = n+i$ for some $i$. $\square$

Corollary 2.2: If $b \geq m+1$, then there exists $a$ such that $n \leq a < m+1$ and $s^b = s^a$. Proof: It suffices to show that there exists $a$ such that $n \leq a < m+1$ and $a \equiv b$ mod $m+1-n$. To that end, let $r$ be the least residue of $b-n$ mod $m+1-n$. That is, $b-n = q(m+1-n) + r$, where $q$ and $r$ are nonnegative. Letting $a = n+r$, we have Then $b = q(m+1-n) + a$ and $n \leq a < m+1$, so $s^b = s^a$. $\square$

Corollary 2.3: $S$ consists of the elements $\{s^1,s^2,\ldots,s^n,\ldots,s^m\}$, and these are pairwise distinct.

Lemma 3: If $a,b \geq n$ and $s^a = s^b$, then $a \equiv b$ mod $m+1-n$. Proof: Since $m+1-n \neq 0$ and $a-n, b-n \geq 0$, there exist nonnegative integers $q_1,q_2,r_1,r_2$ such that $a-n = q_1(m+1-n) + r_1$, $b-n = q_2(m+1-n) + r_2$, and $0 \leq r_1,r_2 < m+1-n$. Since $s^a = s^b$, we have $s^{q_1(m+1-n)+n+r_1} = s^{q_2(m+1-n)+n+r_2}$. Whether $q_1$ and $q_2$ are zero (trivially) or not (Lemma 2), we have $s^{n+r_1} = s^{n+r_2}$. Now $r_1,r_2 < m+1-n$, so that $r_1+n,r_2+n < m+1$. By the minimalness of $m+1$, we have $r_1 + n = r_2 + n$. Thus $a-n \equiv b-n$ mod $m+1-n$, and so $a \equiv b$ mod $m+1-n$. $\square$

Lemma 4: If $a < n$ and $b \in \mathbb{N}^+$ such that $b \neq a$, then $s^a \neq s^b$. Proof: If $b < m+1$ and $s^b = s^a$, we violate the minimalness of $m+1$. If $b \geq m+1$, then by Corollary 2.2, there is an integer $c$ with $n \leq c < m+1$ and $s^c = s^b = s^a$, again violating the minimalness of $m+1$. $\square$

Using Lemmas 3 and 4 and Corollary 2.1, we can now characterize the kernel of $\varphi$ as follows. The equivalence class $[a]$ is $\{a\}$ if $1 \leq a < n$, and is $\{b \geq n \ |\ b \equiv a\ \mathsf{mod}\ m+1-n\}$ if $a \geq n$.

Lemma 5: If $s^a$ is idempotent, with $1 \leq a \leq m$, then $a = d(m+1-n)$. (Where $d$ is as defined in Lemma 1.) Proof: If $1 \leq a < n$, then we have $a = 2a$, which has no solution in $\mathbb{N}^+$. If $n \leq a \leq m$, then $2a \equiv a$ mod $m+1-n$. So $a \equiv 0$ mod $m+1-n$, and then $a = q(m+1-n)$ for some $q$. By Lemma 1, $q = d$. $\square$

So $S$ has precisely one idempotent.

Lemma 6: Let $S$ be a semigroup. Let $E(S)$ denote the set of idempotents in $S$, and let $\mathsf{MG}(S)$ denote the set of maximal subgroups of $S$. Let $\psi : \mathsf{MG}(S) \rightarrow E(S)$ be defined by $G \mapsto 1_G$. Then $\psi$ is bijective. Proof: (Surjective) If $e \in S$ is idempotent, then the set $G_e = \{a \in S\ |\ ea = ae = a, ab = ba = e\ \mathrm{for\ some}\ b \in S\}$ is the $\subseteq$-greatest subgroup of $S$ having identity $e$ by 1.4.11. In particular, $G_e$ is a maximal subgroup, and $\psi(G_e) = e$. If $H \subseteq S$ is a subgroup with identity $e$, then $H \leq G_e$. If $H$ is maximal, then $H = G_e$. So $\psi$ is injective, and thus bijective. $\square$

Since our cyclic semigroup $S$ has only one idempotent, it has a unique maximal subgroup. Namely, $G_e$ where $e = s^{d(m+1-n)}$.

Lemma 7: $s^{d(m+1-n)}s^k = s^k$ if and only if $k \geq n$. Proof: The $(\Leftarrow)$ direction was proved in Lemma 2. Now suppose $s^{d(m+1-n) + k} = s^k$. Now $d(m+1-n) + k \geq n$, so that $k \geq n$ (using our characterization of $\mathsf{ker}\ \varphi$). $\square$

Lemma 8: If $k \geq n$, then there exists $\ell \geq n$ such that $s^ks^\ell = s^{d(m+1-n)}$. Proof: Choose $\ell \geq n$ such that $k+\ell \equiv 0$ mod $d(m+1-n)$. $\square$

Lemmas 7 and 8 demonstrate that $G_e = \{s^t\ |\ n \leq t \leq m\}$.

In particular, $S$ is a group if and only if $n = 1$.

Recall now that the kernel of a semigroup is the intersection of its two sided ideals.

Lemma 9: If $I \subseteq S$ is an ideal, then $G_e \subseteq I$. Proof: Suppose $s^a \in I$. For all $k \geq n$, there exists $b \geq 1$ such that $a+b \geq n$ and $a+b \equiv k$ mod $m+1-n$; then $s^bs^a = s^k$. So $G_e \subseteq I$. $\square$

Finally, using Corollary 2.2 if necessary, we can see that $G_e$ is itself an ideal of $S$. Thus the kernel of $S$ is $G_e$.

Finally, suppose $S$ is a finite semigroup and let $s \in S$. Now $(s)$ is a cyclic subsemigroup of $S$, and so must be finite. As we argued above, $(s)$, and thus $S$, contains an idempotent.