After proving cancellation laws and the idempotent property, we now study equations in groups. One of the strongest consequences of the group axioms is that equations of the forms $a\circ x=b$ and $y\circ a=b$ always have unique solutions. This result shows why groups are algebraic systems where division-like operations are always possible, even when the operation is not ordinary multiplication. In this lesson, students will learn how to solve left and right group equations, why the two formulas differ in non-commutative groups, and how uniqueness follows from cancellation laws. :::definition[Group Equation] Let $(G,\circ)$ be a group and let $a,b\in G$. An equation in which the unknown element belongs to $G$ and is combined with known group elements using $\circ$ is called a $group equation$. Two basic forms are $$\begin{equation*} a\circ x=b \end{equation*}$$ and $$\begin{equation*} y\circ a=b. \end{equation*}$$ ::: In ordinary arithmetic, the equation $ax=b$ is solved by multiplying by $a^{-1}$ when $a\neq0$. In a group, the same idea works, but the side matters. To solve $a\circ x=b$, multiply by $a^{-1}$ on the left. To solve $y\circ a=b$, multiply by $a^{-1}$ on the right. If the group is not commutative, these two operations cannot be interchanged. This side discipline is a major theme in abstract algebra. :::theorem Let $(G,\circ)$ be a group with identity element $e$ and let $a,b\in G$. Then the equation $$\begin{equation*} a\circ x=b \end{equation*}$$ has a unique solution in $G$, namely $$\begin{equation*} x=a^{-1}\circ b. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$ and $a,b\in G$. To prove that the equation $a\circ x=b$ has a unique solution in $G$. Let $$\begin{equation*} x=a^{-1}\circ b. \end{equation*}$$ Since $a^{-1}\in G$, $b\in G$, and $G$ is closed under $\circ$, we get $x\in G$. Now $$\begin{align*} a\circ x &=a\circ(a^{-1}\circ b)\\ &=(a\circ a^{-1})\circ b \quad [\because\, \circ \text{ is associative}]\\ &=e\circ b \quad [\because\, a\circ a^{-1}=e]\\ &=b \quad [\because\, e \text{ is the identity element}]. \end{align*}$$ Therefore $x=a^{-1}\circ b$ is a solution of $a\circ x=b$. Let $x_1,x_2\in G$ be two solutions of $a\circ x=b$. Then $$\begin{equation*} a\circ x_1=b \end{equation*}$$ and $$\begin{equation*} a\circ x_2=b. \end{equation*}$$ Therefore $$\begin{align*} a\circ x_1=b,\quad a\circ x_2=b &\implies a\circ x_1=a\circ x_2\\ &\implies x_1=x_2 \quad [\because\, \text{left cancellation law}]. \end{align*}$$ Hence, $a\circ x=b$ has a unique solution in $G$. $\square$ ::: :::theorem Let $(G,\circ)$ be a group with identity element $e$ and let $a,b\in G$. Then the equation $$\begin{equation*} y\circ a=b \end{equation*}$$ has a unique solution in $G$, namely $$\begin{equation*} y=b\circ a^{-1}. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$ and $a,b\in G$. To prove that the equation $y\circ a=b$ has a unique solution in $G$. Let $$\begin{equation*} y=b\circ a^{-1}. \end{equation*}$$ Since $b\in G$, $a^{-1}\in G$, and $G$ is closed under $\circ$, we get $y\in G$. Now $$\begin{align*} y\circ a &=(b\circ a^{-1})\circ a\\ &=b\circ(a^{-1}\circ a) \quad [\because\, \circ \text{ is associative}]\\ &=b\circ e \quad [\because\, a^{-1}\circ a=e]\\ &=b \quad [\because\, e \text{ is the identity element}]. \end{align*}$$ Therefore $y=b\circ a^{-1}$ is a solution of $y\circ a=b$. Let $y_1,y_2\in G$ be two solutions of $y\circ a=b$. Then $$\begin{equation*} y_1\circ a=b \end{equation*}$$ and $$\begin{equation*} y_2\circ a=b. \end{equation*}$$ Therefore $$\begin{align*} y_1\circ a=b,\quad y_2\circ a=b &\implies y_1\circ a=y_2\circ a\\ &\implies y_1=y_2 \quad [\because\, \text{right cancellation law}]. \end{align*}$$ Hence, $y\circ a=b$ has a unique solution in $G$. $\square$ ::: Group equations are solved by applying the inverse on the correct side. In additive modular arithmetic the operation is commutative, so the numerical answers may look similar, but the formulas still teach the side discipline. Choose an equation type and values, then check the computed solution by substitution. The preview also includes a three-term equation, where the unknown must be isolated by undoing the left factor first and the right factor last. This mirrors the algebraic proof in a finite model. :::scientific-preview[Group Equation Solver Explorer] ::: The two formulas should be memorized with their sides: $$\begin{equation*} a\circ x=b \quad \Rightarrow \quad x=a^{-1}\circ b \end{equation*}$$ and $$\begin{equation*} y\circ a=b \quad \Rightarrow \quad y=b\circ a^{-1}. \end{equation*}$$ In a commutative group, these formulas may look similar, but in a non-commutative group they are genuinely different. Students often write $x=b\circ a^{-1}$ for the first equation. That is generally wrong unless commutativity is known. :::example In the additive group $(\mathbb{Z},+)$, the equation $$\begin{equation*} 5+x=12 \end{equation*}$$ has solution $$\begin{align*} x&=(-5)+12\\ &=7. \end{align*}$$ This matches the group formula $x=a^{-1}\circ b$, because the additive inverse of $5$ is $-5$. ::: :::example In the multiplicative group $(\mathbb{Q}-\{0\},\cdot)$, the equation $$\begin{equation*} 3x=10 \end{equation*}$$ has solution $$\begin{align*} x&=3^{-1}\cdot10\\ &=\frac{1}{3}\cdot10\\ &=\frac{10}{3}. \end{align*}$$ Hence, the solution is $x=\dfrac{10}{3}$. ::: :::solved-problem Let $(G,\circ)$ be a group. Solve the equation $$\begin{equation*} a\circ x\circ b=c \end{equation*}$$ for $x\in G$, where $a,b,c\in G$. ::: :::solution Let $(G,\circ)$ be a group with identity element $e$, and let $a,b,c\in G$. To solve $$\begin{equation*} a\circ x\circ b=c \end{equation*}$$ for $x\in G$. Starting with the equation, $$\begin{align*} a\circ x\circ b=c &\implies a^{-1}\circ(a\circ x\circ b)=a^{-1}\circ c\\ &\implies (a^{-1}\circ a)\circ x\circ b=a^{-1}\circ c \quad [\because\, \circ \text{ is associative}]\\ &\implies e\circ x\circ b=a^{-1}\circ c\\ &\implies x\circ b=a^{-1}\circ c. \end{align*}$$ Now multiply both sides on the right by $b^{-1}$. Then $$\begin{align*} x\circ b=a^{-1}\circ c &\implies (x\circ b)\circ b^{-1}=(a^{-1}\circ c)\circ b^{-1}\\ &\implies x\circ(b\circ b^{-1})=(a^{-1}\circ c)\circ b^{-1}\\ &\implies x\circ e=(a^{-1}\circ c)\circ b^{-1}\\ &\implies x=(a^{-1}\circ c)\circ b^{-1}. \end{align*}$$ Hence, the solution is $$\begin{equation*} x=(a^{-1}\circ c)\circ b^{-1}. \end{equation*}$$ $\square$ ::: Use the calculator below to solve equations in the additive group of integers modulo $n$. Enter $n$, $a$, and $b$. The calculator solves $a+x=b$ modulo $n$ using the group formula $x=-a+b$. This is a finite concrete model of the general group equation $a\circ x=b$. :::calculator[Equation Solver] @libraries: none ::: :::exercise [1] In a group $(G,\circ)$, solve $a\circ x=b$ for $x$. [2] In a group $(G,\circ)$, solve $y\circ a=b$ for $y$. [3] In $(\mathbb{Z},+)$, solve $8+x=-3$. [4] In $(\mathbb{Q}-\{0\},\cdot)$, solve $5x=2$. [5] In a group, solve $p\circ x\circ q=r$. ::: :::answer [1] $x=a^{-1}\circ b$. [2] $y=b\circ a^{-1}$. [3] $x=-11$. [4] $x=\dfrac{2}{5}$. [5] $x=(p^{-1}\circ r)\circ q^{-1}$. ::: :::faq Q: Why does $a\circ x=b$ always have a solution in a group? A: Because $a$ has an inverse, and $x=a^{-1}\circ b$ is in the group. Q: Why is the solution unique? A: If two solutions exist, the left or right cancellation law forces them to be equal. Q: Can $a^{-1}\circ b$ be replaced by $b\circ a^{-1}$? A: Not in general. That replacement requires commutativity. Q: What is the main difference between $a\circ x=b$ and $y\circ a=b$? A: The first equation is solved by multiplying on the left by $a^{-1}$, while the second is solved by multiplying on the right by $a^{-1}$. ::: :::call-to-action[Properties of Inverses] subtitle: Continue to the main inverse laws, including the inverse of the identity, double inverse, and inverse of a product. button: Next Lesson | /abstract-algebra/introduction-to-groups/elementary-properties-of-groups/properties-of-inverses button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/elementary-properties-of-groups/idempotent-elements-in-groups :::