After defining semigroups, we now study a powerful criterion that turns a semigroup into a group. In a group, the equations $a\circ x=b$ and $y\circ a=b$ have unique solutions for all $a,b\in G$. The theorem in this lesson proves a converse: if a semigroup already has this unique solvability property, then it must contain an identity element and inverses, and hence it is a group. This is an important structural result because it shows that the ability to solve equations can replace the explicit assumption of identity and inverse elements. :::definition[Unique Solvability] Let $(S,\circ)$ be a semigroup. The equations $$\begin{equation*} a\circ x=b \end{equation*}$$ and $$\begin{equation*} y\circ a=b \end{equation*}$$ are said to be $uniquely solvable$ in $S$ for all $a,b\in S$ if for every choice of $a,b\in S$, each equation has exactly one solution in $S$. ::: Unique solvability has two parts: existence and uniqueness. Existence means that a solution can always be found inside the semigroup. Uniqueness means that no two different elements solve the same equation. In a group, existence comes from inverses, and uniqueness comes from cancellation. In a semigroup, if both features are assumed for the two basic equations, then identity and inverse elements can be constructed from the equations themselves. :::theorem Let $(S,\circ)$ be a semigroup. If the equations $$\begin{equation*} a\circ x=b \end{equation*}$$ and $$\begin{equation*} y\circ a=b \end{equation*}$$ have unique solutions in $S$ for all $a,b\in S$, then $(S,\circ)$ is a group. ::: :::proof Given that $(S,\circ)$ is a semigroup and the equations $a\circ x=b$ and $y\circ a=b$ have unique solutions in $S$ for all $a,b\in S$. To prove that $(S,\circ)$ is a group. Since $(S,\circ)$ is a semigroup, $\circ$ is associative on $S$. First we prove the existence of a left identity. Choose a fixed element $a\in S$. Since the equation $a\circ x=a$ has a unique solution in $S$, there exists $e_l\in S$ such that $$\begin{equation*} a\circ e_l=a. \end{equation*}$$ Let $u\in S$. Then $a\circ u\in S$. Now $$\begin{align*} a\circ(e_l\circ u) &=(a\circ e_l)\circ u \quad [\because\, \circ \text{ is associative}]\\ &=a\circ u. \end{align*}$$ Also, $$\begin{equation*} a\circ u=a\circ u. \end{equation*}$$ Since the equation $a\circ x=a\circ u$ has a unique solution in $S$, we get $$\begin{equation*} e_l\circ u=u. \end{equation*}$$ Since $u\in S$ is arbitrary, $$\begin{equation*} e_l\circ u=u \quad \forall\, u\in S. \end{equation*}$$ Therefore $e_l$ is a left identity element of $S$. Next we prove the existence of a right identity. Since the equation $y\circ a=a$ has a unique solution in $S$, there exists $e_r\in S$ such that $$\begin{equation*} e_r\circ a=a. \end{equation*}$$ Let $v\in S$. Then $v\circ a\in S$. Now $$\begin{align*} (v\circ e_r)\circ a &=v\circ(e_r\circ a) \quad [\because\, \circ \text{ is associative}]\\ &=v\circ a. \end{align*}$$ Also, $$\begin{equation*} v\circ a=v\circ a. \end{equation*}$$ Since the equation $y\circ a=v\circ a$ has a unique solution in $S$, we get $$\begin{equation*} v\circ e_r=v. \end{equation*}$$ Since $v\in S$ is arbitrary, $$\begin{equation*} v\circ e_r=v \quad \forall\, v\in S. \end{equation*}$$ Therefore $e_r$ is a right identity element of $S$. Now $$\begin{align*} e_l\circ e_r&=e_r \quad [\because\, e_l \text{ is a left identity element}]\\ e_l\circ e_r&=e_l \quad [\because\, e_r \text{ is a right identity element}]. \end{align*}$$ Therefore $$\begin{equation*} e_l=e_r. \end{equation*}$$ Let $$\begin{equation*} e=e_l=e_r. \end{equation*}$$ Then $$\begin{equation*} e\circ s=s\circ e=s \quad \forall\, s\in S. \end{equation*}$$ Therefore $e$ is an identity element of $S$. It remains to prove the existence of inverse elements. Let $g\in S$. Since the equation $g\circ x=e$ has a unique solution in $S$, there exists $p\in S$ such that $$\begin{equation*} g\circ p=e. \end{equation*}$$ Since the equation $y\circ g=e$ has a unique solution in $S$, there exists $q\in S$ such that $$\begin{equation*} q\circ g=e. \end{equation*}$$ Now $$\begin{align*} p &=e\circ p \quad [\because\, e \text{ is the identity element}]\\ &=(q\circ g)\circ p \quad [\because\, q\circ g=e]\\ &=q\circ(g\circ p) \quad [\because\, \circ \text{ is associative}]\\ &=q\circ e \quad [\because\, g\circ p=e]\\ &=q \quad [\because\, e \text{ is the identity element}]. \end{align*}$$ Therefore $p=q$. Thus for each $g\in S$, there exists $p\in S$ such that $$\begin{equation*} g\circ p=p\circ g=e. \end{equation*}$$ Therefore every element of $S$ has an inverse in $S$. Hence, $(S,\circ)$ is a group. $\square$ ::: The proof has three conceptual stages. First, a left identity is constructed from the equation $a\circ x=a$. Second, a right identity is constructed from the equation $y\circ a=a$. Third, the two identities are shown to be equal, and then inverse elements are obtained by solving equations with the new identity element. The proof depends critically on associativity, so the semigroup hypothesis cannot be dropped. :::example Let $(G,\circ)$ be a group. Then the equations $a\circ x=b$ and $y\circ a=b$ are uniquely solvable for all $a,b\in G$. Indeed, the solutions are $$\begin{equation*} x=a^{-1}\circ b \end{equation*}$$ and $$\begin{equation*} y=b\circ a^{-1}. \end{equation*}$$ Thus groups satisfy the unique solvability condition. The theorem proves that, among semigroups, this condition is also sufficient for being a group. ::: :::solved-problem Let $(S,\circ)$ be a semigroup in which the equations $a\circ x=b$ and $y\circ a=b$ have unique solutions for all $a,b\in S$. Prove that $S$ has an identity element. ::: :::solution Let $(S,\circ)$ be a semigroup in which the equations $a\circ x=b$ and $y\circ a=b$ have unique solutions for all $a,b\in S$. To prove that $S$ has an identity element. Choose $a\in S$. Since $a\circ x=a$ has a unique solution, there exists $e_l\in S$ such that $$\begin{equation*} a\circ e_l=a. \end{equation*}$$ Let $u\in S$. Then $$\begin{align*} a\circ(e_l\circ u) &=(a\circ e_l)\circ u\\ &=a\circ u. \end{align*}$$ By uniqueness of the solution of $a\circ x=a\circ u$, we get $$\begin{equation*} e_l\circ u=u \quad \forall\, u\in S. \end{equation*}$$ Thus $e_l$ is a left identity element. Similarly, since $y\circ a=a$ has a unique solution, there exists $e_r\in S$ such that $$\begin{equation*} e_r\circ a=a. \end{equation*}$$ Let $v\in S$. Then $$\begin{align*} (v\circ e_r)\circ a &=v\circ(e_r\circ a)\\ &=v\circ a. \end{align*}$$ By uniqueness of the solution of $y\circ a=v\circ a$, we get $$\begin{equation*} v\circ e_r=v \quad \forall\, v\in S. \end{equation*}$$ Thus $e_r$ is a right identity element. Now $$\begin{align*} e_l\circ e_r&=e_r\\ e_l\circ e_r&=e_l. \end{align*}$$ Therefore $e_l=e_r$. Hence, $S$ has an identity element. $\square$ ::: :::remark In the first part of the proof, it is important to let $u\in S$ be arbitrary. This corrects a common mistake: one cannot conclude that $e_l$ is a left identity merely from $a\circ e_l=a$ for one fixed element $a$. The argument must show that $e_l\circ u=u$ for every $u\in S$. The unique solvability hypothesis is exactly what allows this extension from one equation to all elements. ::: Unique solvability can be checked by counting solutions to left and right equations. We observe addition modulo $n$, where each equation $a+x=b$ has exactly one solution and each equation $y+a=b$ has exactly one solution. Change the modulus and the selected values of $a$ and $b$ to see the unique solution directly. This finite model shows the equation property that the theorem abstracts to arbitrary semigroups. :::scientific-preview[Equation Solution Counts] @libraries: plotly ::: Use the calculator for a finite set $\{0,1,\ldots,n-1\}$ under addition modulo $n$. It checks whether every equation $a+x=b$ and $y+a=b$ has a unique solution. This finite example illustrates the equation criterion in a familiar group. :::calculator[Equation Criterion Test] ::: :::exercise [1] State the unique solvability criterion for a semigroup to be a group. [2] Explain why associativity is needed in the proof. [3] In a group, solve $a\circ x=b$ and $y\circ a=b$. [4] In the proof of the criterion, why must the left identity be shown to work for every element? [5] If a semigroup has unique solutions for $a\circ x=b$ but not necessarily for $y\circ a=b$, can the theorem be applied? ::: :::answer [1] If $(S,\circ)$ is a semigroup and both equations $a\circ x=b$ and $y\circ a=b$ have unique solutions in $S$ for all $a,b\in S$, then $(S,\circ)$ is a group. [2] Associativity is used to rewrite expressions such as $a\circ(e_l\circ u)$ as $(a\circ e_l)\circ u$. [3] The solutions are $x=a^{-1}\circ b$ and $y=b\circ a^{-1}$. [4] An identity element must act on every element of the set, not only on one fixed element. [5] No. The theorem requires unique solvability of both left and right equations. ::: :::faq Q: What does unique solvability mean? A: It means that each equation has exactly one solution inside the semigroup. Q: Why does the theorem start with a semigroup? A: The proof uses associativity, which is the defining law of a semigroup. Q: Does unique solvability give inverses? A: Yes. After constructing the identity element, the equations $g\circ x=e$ and $y\circ g=e$ give inverse elements. Q: Is this theorem a converse of the group equation theorem? A: Yes. Groups have uniquely solvable equations, and this theorem shows that a semigroup with those uniquely solvable equations is a group. ::: :::call-to-action[Finite Cancellative Semigroups] subtitle: Continue to finite cancellative semigroups and see how cancellation laws force a finite semigroup to become a group. button: Next Lesson | /abstract-algebra/introduction-to-groups/semigroup-conditions-for-groups/finite-cancellative-semigroups button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/semigroup-conditions-for-groups/semigroups :::