After proving that identity elements and inverse elements are unique, we now prove the cancellation laws in a group. Cancellation is one of the most useful algebraic tools: it allows us to remove a common factor from the left or from the right of an equation. In ordinary arithmetic, students cancel numbers almost automatically, but in abstract algebra cancellation must be justified from the group axioms. The proof uses inverses and associativity. In this lesson, students will learn the left cancellation law, the right cancellation law, and the correct way to apply them in group equations. :::definition[Left Cancellation Law] Let $(G,\circ)$ be a group. The $left cancellation law$ says that for all $a,b,c\in G$, $$\begin{equation*} a\circ b=a\circ c \implies b=c. \end{equation*}$$ It means that a common left factor may be cancelled from both sides of a group equation. ::: :::definition[Right Cancellation Law] Let $(G,\circ)$ be a group. The $right cancellation law$ says that for all $a,b,c\in G$, $$\begin{equation*} b\circ a=c\circ a \implies b=c. \end{equation*}$$ It means that a common right factor may be cancelled from both sides of a group equation. ::: The two cancellation laws must be stated separately because a general group need not be commutative. If $a\circ b=a\circ c$, then $a$ is a common left factor. To cancel it, we multiply both sides on the left by $a^{-1}$. If $b\circ a=c\circ a$, then $a$ is a common right factor. To cancel it, we multiply both sides on the right by $a^{-1}$. A common mistake is to multiply on the wrong side in a non-commutative setting. :::theorem Let $(G,\circ)$ be a group with identity element $e$. If $a,b,c\in G$, then the cancellation laws hold in $G$, that is, $$\begin{equation*} a\circ b=a\circ c \implies b=c \end{equation*}$$ and $$\begin{equation*} b\circ a=c\circ a \implies b=c. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$ and $a,b,c\in G$. To prove that the cancellation laws hold in $G$. [1] To prove that $a\circ b=a\circ c \implies b=c$. Let $$\begin{equation*} a\circ b=a\circ c. \end{equation*}$$ Then $$\begin{align*} a\circ b=a\circ c &\implies a^{-1}\circ(a\circ b)=a^{-1}\circ(a\circ c)\\ &\implies (a^{-1}\circ a)\circ b=(a^{-1}\circ a)\circ c \quad [\because\, \circ \text{ is associative}]\\ &\implies e\circ b=e\circ c \quad [\because\, a^{-1}\circ a=e]\\ &\implies b=c \quad [\because\, e \text{ is the identity element}]. \end{align*}$$ Therefore $$\begin{equation*} a\circ b=a\circ c \implies b=c. \end{equation*}$$ [2] To prove that $b\circ a=c\circ a \implies b=c$. Let $$\begin{equation*} b\circ a=c\circ a. \end{equation*}$$ Then $$\begin{align*} b\circ a=c\circ a &\implies (b\circ a)\circ a^{-1}=(c\circ a)\circ a^{-1}\\ &\implies b\circ(a\circ a^{-1})=c\circ(a\circ a^{-1}) \quad [\because\, \circ \text{ is associative}]\\ &\implies b\circ e=c\circ e \quad [\because\, a\circ a^{-1}=e]\\ &\implies b=c \quad [\because\, e \text{ is the identity element}]. \end{align*}$$ Hence, the cancellation laws hold in $G$. $\square$ ::: Cancellation works in every group because every element has an inverse. When the inverse condition fails, a common factor can collapse different elements to the same result. Choose a modulus, an operation, and a common factor to search for such collapses. The preview compares left cancellation and right cancellation separately. Addition modulo $n$ behaves like a group on all residues, while multiplication modulo $n$ on all residues often exposes cancellation failures. :::scientific-preview[Cancellation Law Explorer] ::: This theorem explains why cancellation works in every group. It does not depend on numbers, order, or commutativity. The essential reason is the existence of inverses. A common left factor is removed by applying the inverse of that factor on the left. A common right factor is removed by applying the inverse on the right. This side discipline is central in matrix groups, permutation groups, and other non-commutative examples. :::example In the additive group $(\mathbb{Z},+)$, the left and right cancellation laws become the usual cancellation law: $$\begin{equation*} a+b=a+c \implies b=c. \end{equation*}$$ Indeed, adding $-a$ to both sides gives $$\begin{align*} a+b=a+c &\implies (-a)+(a+b)=(-a)+(a+c)\\ &\implies ((-a)+a)+b=((-a)+a)+c\\ &\implies 0+b=0+c\\ &\implies b=c. \end{align*}$$ Hence, cancellation in integers under addition is a special case of group cancellation. ::: :::example In the multiplicative group $(\mathbb{R}-\{0\},\cdot)$, $$\begin{equation*} ab=ac \implies b=c. \end{equation*}$$ Here $a\neq0$, so $a^{-1}=\dfrac{1}{a}$ exists. Multiplying both sides by $\dfrac{1}{a}$ gives $$\begin{align*} ab=ac &\implies \frac{1}{a}(ab)=\frac{1}{a}(ac)\\ &\implies \left(\frac{1}{a}a\right)b=\left(\frac{1}{a}a\right)c\\ &\implies 1b=1c\\ &\implies b=c. \end{align*}$$ Hence, ordinary nonzero real cancellation follows from the group law. ::: :::solved-problem Let $(G,\circ)$ be a group and let $a,b\in G$. If $$\begin{equation*} a\circ b=a, \end{equation*}$$ prove that $b=e$. ::: :::solution Let $(G,\circ)$ be a group with identity element $e$, and let $a,b\in G$. Given that $$\begin{equation*} a\circ b=a. \end{equation*}$$ To prove that $b=e$. Since $e$ is the identity element, $$\begin{equation*} a\circ e=a. \end{equation*}$$ Therefore $$\begin{align*} a\circ b=a &\implies a\circ b=a\circ e \quad [\because\, a\circ e=a]\\ &\implies b=e \quad [\because\, \text{left cancellation law}]. \end{align*}$$ Hence, $b=e$. $\square$ ::: :::solved-problem Let $(G,\circ)$ be a group and let $a,b\in G$. If $$\begin{equation*} b\circ a=a, \end{equation*}$$ prove that $b=e$. ::: :::solution Let $(G,\circ)$ be a group with identity element $e$, and let $a,b\in G$. Given that $$\begin{equation*} b\circ a=a. \end{equation*}$$ To prove that $b=e$. Since $e$ is the identity element, $$\begin{equation*} e\circ a=a. \end{equation*}$$ Therefore $$\begin{align*} b\circ a=a &\implies b\circ a=e\circ a \quad [\because\, e\circ a=a]\\ &\implies b=e \quad [\because\, \text{right cancellation law}]. \end{align*}$$ Hence, $b=e$. $\square$ ::: :::remark Cancellation is not a valid law in every algebraic structure. For example, in ordinary multiplication on $\mathbb{Z}$, the equation $0\cdot2=0\cdot3$ is true, but $2\neq3$. The failure occurs because $0$ has no multiplicative inverse in $\mathbb{Z}$. Thus cancellation in groups is not a habit; it is a theorem supported by inverses. ::: Use the calculator below to compare cancellation in modular arithmetic. Choose a modulus $n$ and a common factor $a$. The calculator searches for a case where $a+b$ and $a+c$ agree modulo $n$ but $b$ and $c$ differ, or where $ab$ and $ac$ agree modulo $n$ but $b$ and $c$ differ. This helps students see why cancellation is safe only when the group hypotheses are satisfied. :::calculator[Cancellation Test] @libraries: none ::: :::exercise [1] Prove that if $a\circ b=a\circ c$ in a group, then $b=c$. [2] Prove that if $b\circ a=c\circ a$ in a group, then $b=c$. [3] Let $(G,\circ)$ be a group. If $a\circ b=e\circ b$, prove that $a=e$. [4] Let $(G,\circ)$ be a group. If $a\circ b=a\circ e$, prove that $b=e$. [5] Give an example from ordinary arithmetic where cancellation fails outside a group. ::: :::answer [1] Multiply both sides on the left by $a^{-1}$ and use associativity. [2] Multiply both sides on the right by $a^{-1}$ and use associativity. [3] Apply the right cancellation law to $a\circ b=e\circ b$. [4] Apply the left cancellation law to $a\circ b=a\circ e$. [5] In $\mathbb{Z}$ under multiplication, $0\cdot2=0\cdot3$, but $2\neq3$. ::: :::faq Q: Why are there two cancellation laws? A: In a non-commutative group, cancelling a left factor and cancelling a right factor are different operations. Q: Can we cancel any element in a group? A: Yes. Every element of a group has an inverse, so every element can be cancelled from the correct side. Q: Does cancellation require commutativity? A: No. Cancellation holds in every group, including non-commutative groups. Q: Why can cancellation fail outside groups? A: Cancellation can fail when the common factor has no inverse. ::: :::call-to-action[Idempotent Element] subtitle: Continue to idempotent elements and learn why the identity is the only idempotent element in a group. button: Next Lesson | /abstract-algebra/introduction-to-groups/elementary-properties-of-groups/idempotent-elements-in-groups button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/elementary-properties-of-groups/inverse-element-in-groups :::