After centre and centralisers, we now pass from special subgroups to the way a subgroup sits inside the whole group. The main idea in this lecture is the coset of a subgroup. A coset is obtained by multiplying every element of a subgroup by one fixed element of the group. This construction matters because it is the first systematic method for dividing a group into equal-sized parts. In later lectures, cosets will lead to index, Lagrange's theorem, quotient groups, and normal subgroups. In this class note, we define left and right cosets, compute first examples, and see why nonabelian groups require us to distinguish the two sides. :::definition[Left Coset] Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. For $a\in G$, the $left coset$ of $H$ in $G$ determined by $a$ is the subset $$\begin{equation*} aH=\{a\circ h:h\in H\}. \end{equation*}$$ The element $a$ is called a representative of the left coset $aH$. ::: :::definition[Right Coset] Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. For $a\in G$, the $right coset$ of $H$ in $G$ determined by $a$ is the subset $$\begin{equation*} Ha=\{h\circ a:h\in H\}. \end{equation*}$$ The element $a$ is called a representative of the right coset $Ha$. ::: The definitions differ only in the side on which $a$ is placed. In an abelian group, this difference disappears because $a\circ h=h\circ a$ for every $h\in H$. In a nonabelian group, the left coset $aH$ and the right coset $Ha$ may be different. A common mistake here is to write $aH=Ha$ merely because the same letters appear. That equality needs a reason; it is not automatic in an arbitrary group. We observe left and right cosets first in the additive group of integers modulo $n$. Choose a modulus, a subgroup step, and a representative, then compare the translated set with the original subgroup. In this abelian setting the left and right cosets agree, so the display helps isolate what the definition means before nonabelian complications appear. Try the example $n=6$, subgroup step $3$, and representative $1$ to see the coset $\{1,4\}$. :::scientific-preview[Left and Right Coset Builder] ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $a\in G$, then $$\begin{equation*} a\in aH \end{equation*}$$ and $$\begin{equation*} a\in Ha. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $a\in G$. To prove that $a\in aH$ and $a\in Ha$. Since $H$ is a subgroup of $G$, the identity element $e$ of $G$ belongs to $H$. Now $$\begin{align*} a &=a\circ e &&[\because e \text{ is the identity element}]\\ &\in aH &&[\because e\in H]. \end{align*}$$ Therefore $a\in aH$. Also, $$\begin{align*} a &=e\circ a &&[\because e \text{ is the identity element}]\\ &\in Ha &&[\because e\in H]. \end{align*}$$ Hence, $a\in aH$ and $a\in Ha$. $\square$ ::: :::example[Cosets in an Additive Group] Let $G=\mathbb{Z}_6$ under addition modulo $6$ and let $$\begin{equation*} H=\{0,3\}. \end{equation*}$$ Then $H$ is a subgroup of $G$. Since the operation is addition, the left and right cosets determined by $1$ are both written as $1+H$: $$\begin{align*} 1+H &=\{1+0,\;1+3\}\\ &=\{1,\;4\}. \end{align*}$$ Similarly, $$\begin{align*} 2+H &=\{2+0,\;2+3\}\\ &=\{2,\;5\}. \end{align*}$$ Here left cosets and right cosets are the same because $\mathbb{Z}_6$ under addition is abelian. ::: :::example[Left and Right Cosets May Differ] Let $G=S_3$ be the group of permutations of $\{1,2,3\}$ under composition and let $$\begin{equation*} H=\{e,(12)\}. \end{equation*}$$ Let $a=(123)$. Using composition from right to left, we get $$\begin{align*} aH &=\{(123)e,\;(123)(12)\}\\ &=\{(123),\;(13)\}. \end{align*}$$ On the other hand, $$\begin{align*} Ha &=\{e(123),\;(12)(123)\}\\ &=\{(123),\;(23)\}. \end{align*}$$ Therefore $$\begin{equation*} aH\neq Ha. \end{equation*}$$ Hence, the side on which the representative is multiplied matters in a nonabelian group. ::: :::solved-problem[Computing a Coset] Let $G=\mathbb{Z}_{12}$ under addition modulo $12$ and let $H=\{0,4,8\}$. Find $5+H$. ::: :::solution[Solution] Let $G=\mathbb{Z}_{12}$ and $H=\{0,4,8\}$. Then $$\begin{align*} 5+H &=\{5+0,\;5+4,\;5+8\}\\ &=\{5,\;9,\;13\}\\ &=\{5,\;9,\;1\} &&[\because 13\equiv 1 \pmod {12}]. \end{align*}$$ Therefore $$\begin{equation*} 5+H=\{1,5,9\}. \end{equation*}$$ ::: We compute cosets in $\mathbb{Z}_n$ by adding the same representative to every subgroup element and reducing modulo $n$. Enter a modulus, subgroup step, and representative to list the subgroup and the translated coset. The calculator sorts the residues so that repeated representatives are easy to compare. This reinforces that a coset is a set, so the order in which residues are listed does not matter. :::calculator[Modular Coset Calculator] ::: :::exercise[Exercises] 1. Let $G=\mathbb{Z}_{10}$ under addition modulo $10$ and let $H=\{0,5\}$. Find $2+H$. 2. Let $G=\mathbb{Z}_{8}$ under addition modulo $8$ and let $H=\{0,4\}$. Find $3+H$. 3. Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Prove that $eH=H$ and $He=H$. ::: :::answer[Answers] 1. We have $$\begin{equation*} 2+H=\{2,7\}. \end{equation*}$$ 2. We have $$\begin{equation*} 3+H=\{3,7\}. \end{equation*}$$ 3. Since $e\circ h=h$ and $h\circ e=h$ for every $h\in H$, we get $$\begin{equation*} eH=H \end{equation*}$$ and $$\begin{equation*} He=H. \end{equation*}$$ ::: :::faq Q: Is a coset always a subgroup? A: No. A coset need not contain the identity element. For example, in $\mathbb{Z}_6$, the coset $\{1,4\}$ is not a subgroup. Q: Why do we define both left and right cosets? A: In nonabelian groups, $a\circ h$ and $h\circ a$ may be different. Therefore $aH$ and $Ha$ may be different subsets. Q: Can the representative of a coset belong to the coset? A: Yes. Since $e\in H$, we have $a=a\circ e\in aH$ and $a=e\circ a\in Ha$. ::: :::call-to-action subtitle: Continue with the condition under which a coset is exactly the subgroup itself. button: Next: Cosets Determined by Elements of a Subgroup | published/abstract-algebra/introduction-to-groups/cosets-of-subgroups/cosets-determined-by-elements-of-a-subgroup.md :::