Cosets can also be understood through equivalence relations. This viewpoint explains why cosets form partitions: every equivalence relation partitions a set into equivalence classes. For a subgroup $H$ of a group $G$, the relation $a\rho b$ defined by $a^{-1}\circ b\in H$ produces left cosets. A related relation defined by $b\circ a^{-1}\in H$ produces right cosets. This lecture connects the algebraic definition of a coset with the general set-theoretic idea of equivalence classes. :::definition[Left Coset Relation] Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. The $left coset relation$ on $G$ determined by $H$ is the relation $\rho_L$ defined by $$\begin{equation*} a\rho_L b \iff a^{-1}\circ b\in H \end{equation*}$$ for all $a,b\in G$. ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Let $\rho_L$ be the relation on $G$ defined by $$\begin{equation*} a\rho_L b \iff a^{-1}\circ b\in H. \end{equation*}$$ Then $\rho_L$ is an equivalence relation and $$\begin{equation*} \operatorname{cl}(a)=aH \end{equation*}$$ for every $a\in G$. ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $\rho_L$ is defined on $G$ by $$\begin{equation*} a\rho_L b \iff a^{-1}\circ b\in H. \end{equation*}$$ To prove that $\rho_L$ is an equivalence relation and $\operatorname{cl}(a)=aH$ for every $a\in G$. [1] To prove reflexivity. Let $a\in G$. Then $$\begin{align*} a^{-1}\circ a &=e\\ &\in H. \end{align*}$$ Therefore $a\rho_L a$. [2] To prove symmetry. Let $a,b\in G$ and let $a\rho_L b$. Then $$\begin{equation*} a^{-1}\circ b\in H. \end{equation*}$$ Since $H$ is a subgroup, it is closed under inverses. Therefore $$\begin{align*} a^{-1}\circ b\in H &\implies (a^{-1}\circ b)^{-1}\in H\\ &\implies b^{-1}\circ a\in H. \end{align*}$$ Therefore $b\rho_L a$. [3] To prove transitivity. Let $a,b,c\in G$ such that $a\rho_L b$ and $b\rho_L c$. Then $$\begin{equation*} a^{-1}\circ b\in H \end{equation*}$$ and $$\begin{equation*} b^{-1}\circ c\in H. \end{equation*}$$ Since $H$ is closed under $\circ$, we get $$\begin{align*} (a^{-1}\circ b)\circ (b^{-1}\circ c)\in H &\implies a^{-1}\circ c\in H. \end{align*}$$ Therefore $a\rho_L c$. Therefore $\rho_L$ is an equivalence relation. Now let $a\in G$. Then $$\begin{align*} \operatorname{cl}(a) &=\{b\in G:a\rho_L b\}\\ &=\{b\in G:a^{-1}\circ b\in H\}\\ &=\{b\in G:b\in aH\}\\ &=aH. \end{align*}$$ Hence, $\rho_L$ is an equivalence relation and $\operatorname{cl}(a)=aH$ for every $a\in G$. $\square$ ::: We observe the equivalence class of an element under the left coset relation in $\mathbb{Z}_n$. Choose an element $a$ and the preview finds all $b$ such that $-a+b$ lies in $H$. The resulting class is exactly the coset $a+H$. This connects the algebraic translation picture with the set-theoretic idea of equivalence classes. :::scientific-preview[Coset Equivalence Class Explorer] ::: :::definition[Right Coset Relation] Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. The $right coset relation$ on $G$ determined by $H$ is the relation $\rho_R$ defined by $$\begin{equation*} a\rho_R b \iff b\circ a^{-1}\in H \end{equation*}$$ for all $a,b\in G$. ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Let $\rho_R$ be the relation on $G$ defined by $$\begin{equation*} a\rho_R b \iff b\circ a^{-1}\in H. \end{equation*}$$ Then $\rho_R$ is an equivalence relation and $$\begin{equation*} \operatorname{cl}(a)=Ha \end{equation*}$$ for every $a\in G$. ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $\rho_R$ is defined on $G$ by $$\begin{equation*} a\rho_R b \iff b\circ a^{-1}\in H. \end{equation*}$$ To prove that $\rho_R$ is an equivalence relation and $\operatorname{cl}(a)=Ha$ for every $a\in G$. [1] To prove reflexivity. Let $a\in G$. Then $$\begin{align*} a\circ a^{-1} &=e\\ &\in H. \end{align*}$$ Therefore $a\rho_R a$. [2] To prove symmetry. Let $a,b\in G$ and let $a\rho_R b$. Then $$\begin{equation*} b\circ a^{-1}\in H. \end{equation*}$$ Since $H$ is a subgroup, it is closed under inverses. Therefore $$\begin{align*} b\circ a^{-1}\in H &\implies (b\circ a^{-1})^{-1}\in H\\ &\implies a\circ b^{-1}\in H. \end{align*}$$ Therefore $b\rho_R a$. [3] To prove transitivity. Let $a,b,c\in G$ such that $a\rho_R b$ and $b\rho_R c$. Then $$\begin{equation*} b\circ a^{-1}\in H \end{equation*}$$ and $$\begin{equation*} c\circ b^{-1}\in H. \end{equation*}$$ Since $H$ is closed under $\circ$, we get $$\begin{align*} (c\circ b^{-1})\circ (b\circ a^{-1})\in H &\implies c\circ a^{-1}\in H. \end{align*}$$ Therefore $a\rho_R c$. Therefore $\rho_R$ is an equivalence relation. Now let $a\in G$. Then $$\begin{align*} \operatorname{cl}(a) &=\{b\in G:a\rho_R b\}\\ &=\{b\in G:b\circ a^{-1}\in H\}\\ &=\{b\in G:b\in Ha\}\\ &=Ha. \end{align*}$$ Hence, $\rho_R$ is an equivalence relation and $\operatorname{cl}(a)=Ha$ for every $a\in G$. $\square$ ::: :::example[Equivalence Classes in Integers Modulo Twelve] Let $G=\mathbb{Z}_{12}$ under addition modulo $12$ and let $$\begin{equation*} H=\{0,4,8\}. \end{equation*}$$ For the left coset relation, in additive notation, $$\begin{equation*} a\rho_L b \iff -a+b\in H. \end{equation*}$$ The class of $1$ is $$\begin{align*} \operatorname{cl}(1) &=1+H\\ &=\{1,5,9\}. \end{align*}$$ Thus the equivalence class of $1$ is exactly the coset determined by $1$. ::: :::solved-problem[Finding an Equivalence Class] Let $G=\mathbb{Z}_{10}$ under addition modulo $10$ and let $H=\{0,5\}$. For the relation $a\rho b$ if and only if $-a+b\in H$, find $\operatorname{cl}(3)$. ::: :::solution[Solution] Let $G=\mathbb{Z}_{10}$ and $H=\{0,5\}$. The relation is the left coset relation in additive notation. Therefore $$\begin{align*} \operatorname{cl}(3) &=3+H\\ &=\{3+0,\;3+5\}\\ &=\{3,8\}. \end{align*}$$ ::: We compute the equivalence class of an element for the modular left coset relation. The calculator checks each residue $b$ and keeps exactly those satisfying $-a+b\in H$. It then compares the class with the coset $a+H$. This makes the identity between equivalence classes and cosets visible element by element. :::calculator[Equivalence Class Calculator] ::: :::exercise[Exercises] 1. Let $G=\mathbb{Z}_{9}$ and let $H=\{0,3,6\}$. For the relation $a\rho b$ if and only if $-a+b\in H$, find $\operatorname{cl}(2)$. 2. Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Prove that $a\rho_L b$ if and only if $b\in aH$. 3. Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Prove that $a\rho_R b$ if and only if $b\in Ha$. ::: :::answer[Answers] 1. We have $$\begin{align*} \operatorname{cl}(2) &=2+H\\ &=\{2,5,8\}. \end{align*}$$ 2. By definition, $a\rho_L b$ if and only if $a^{-1}\circ b\in H$. By the left coset membership criterion, this is equivalent to $b\in aH$. 3. By definition, $a\rho_R b$ if and only if $b\circ a^{-1}\in H$. By the right coset membership criterion, this is equivalent to $b\in Ha$. ::: :::faq Q: Why do equivalence relations appear in the study of cosets? A: Cosets are equivalence classes for natural relations determined by the subgroup. Q: Which relation gives left cosets? A: The relation $a\rho_L b$ if and only if $a^{-1}\circ b\in H$ gives left cosets. Q: Which relation gives right cosets? A: The relation $a\rho_R b$ if and only if $b\circ a^{-1}\in H$ gives right cosets. ::: :::call-to-action subtitle: Continue with the size of cosets and the correspondence between left and right cosets. button: Next: Cardinality and Correspondence of Cosets | published/abstract-algebra/introduction-to-groups/cosets-of-subgroups/cardinality-and-correspondence-of-cosets.md :::