Having studied normal subgroups, we now use them to build new groups from old groups. The quotient group is formed by taking all cosets of a normal subgroup and multiplying those cosets as if they were elements. This construction is one of the central ideas of abstract algebra because it allows us to collapse a subgroup to the identity and study the remaining structure. Students often remember the formula $(aH)(bH)=(a\circ b)H$ but forget the real issue: this formula must be independent of the chosen representatives. In this lesson, quotient group construction is developed carefully from coset multiplication to the group axioms. :::definition[Quotient Set] Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. The quotient set of $G$ by $H$ is the set of all left cosets of $H$ in $G$, denoted by $$\begin{equation*} G/H=\{aH:a\in G\}. \end{equation*}$$ ::: The notation $G/H$ first denotes a set of cosets. It becomes a group only after a suitable operation is defined on these cosets. The normality of $H$ is what makes the operation behave correctly. Without normality, different representatives of the same coset may give different products, and the construction fails. :::theorem Let $(G,\circ)$ be a group and let $H$ be a normal subgroup of $G$. Define a binary operation $*$ on $G/H$ by $$\begin{equation*} aH*bH=(a\circ b)H. \end{equation*}$$ Then $(G/H,*)$ is a group. ::: :::proof Given that $(G,\circ)$ is a group and $H$ is a normal subgroup of $G$. To prove that $(G/H,*)$ is a group. [1] To prove that $*$ is well-defined. Let $$\begin{equation*} aH=a'H \end{equation*}$$ and $$\begin{equation*} bH=b'H. \end{equation*}$$ Then there exist $h_1,h_2\in H$ such that $$\begin{equation*} a'=a\circ h_1 \end{equation*}$$ and $$\begin{equation*} b'=b\circ h_2. \end{equation*}$$ Now $$\begin{align*} (a\circ b)^{-1}\circ(a'\circ b') &=b^{-1}\circ a^{-1}\circ(a\circ h_1)\circ(b\circ h_2)\\ &=b^{-1}\circ h_1\circ b\circ h_2. \end{align*}$$ Since $H$ is normal in $G$ and $h_1\in H$, we get $$\begin{equation*} b^{-1}\circ h_1\circ b\in H. \end{equation*}$$ Since $h_2\in H$, we get $$\begin{equation*} (a\circ b)^{-1}\circ(a'\circ b')\in H. \end{equation*}$$ Therefore $$\begin{equation*} (a\circ b)H=(a'\circ b')H. \end{equation*}$$ Thus $*$ is well-defined. [2] To prove closure. Let $aH,bH\in G/H$. Then $a\circ b\in G$, and therefore $$\begin{equation*} aH*bH=(a\circ b)H\in G/H. \end{equation*}$$ [3] To prove associativity. Let $aH,bH,cH\in G/H$. Then $$\begin{align*} (aH*bH)*cH &=((a\circ b)H)*cH\\ &=((a\circ b)\circ c)H\\ &=(a\circ(b\circ c))H\\ &=aH*((b\circ c)H)\\ &=aH*(bH*cH). \end{align*}$$ Therefore $*$ is associative. [4] To prove the existence of an identity element. The identity element is $H=eH$. For every $aH\in G/H$, $$\begin{align*} aH*H &=aH*eH\\ &=(a\circ e)H\\ &=aH \end{align*}$$ and $$\begin{align*} H*aH &=eH*aH\\ &=(e\circ a)H\\ &=aH. \end{align*}$$ Therefore $H$ is the identity element of $G/H$. [5] To prove the existence of inverses. Let $aH\in G/H$. Then $a^{-1}H\in G/H$, and $$\begin{align*} aH*a^{-1}H &=(a\circ a^{-1})H\\ &=eH\\ &=H \end{align*}$$ and $$\begin{align*} a^{-1}H*aH &=(a^{-1}\circ a)H\\ &=eH\\ &=H. \end{align*}$$ Therefore every element of $G/H$ has an inverse in $G/H$. Hence, $(G/H,*)$ is a group. $\square$ ::: :::definition[Quotient Group] Let $(G,\circ)$ be a group and let $H$ be a normal subgroup of $G$. The group $$\begin{equation*} G/H=\{aH:a\in G\} \end{equation*}$$ under the operation $$\begin{equation*} aH*bH=(a\circ b)H \end{equation*}$$ is called the quotient group or factor group of $G$ by $H$. ::: :::example Let $G=(\mathbb{Z},+)$ and let $H=3\mathbb{Z}$. Since $\mathbb{Z}$ is abelian, $3\mathbb{Z}$ is normal in $\mathbb{Z}$. The quotient group is $$\begin{equation*} \mathbb{Z}/3\mathbb{Z}=\{0+3\mathbb{Z},1+3\mathbb{Z},2+3\mathbb{Z}\}. \end{equation*}$$ The operation is $$\begin{equation*} (a+3\mathbb{Z})+(b+3\mathbb{Z})=(a+b)+3\mathbb{Z}. \end{equation*}$$ Hence, $\mathbb{Z}/3\mathbb{Z}$ is a quotient group. $\square$ ::: We observe quotient multiplication in $\mathbb{Z}/n\mathbb{Z}$. Choose two representatives and a modulus, then compare the representative sum with the resulting coset. The important idea is that many different representatives can name the same coset, but the product rule still gives the same answer. This is the well-defined behaviour that normality guarantees in the general construction. :::scientific-preview[Quotient Coset Multiplication Explorer] ::: A quotient element is a whole coset, not a single integer. Changing a representative by an element of the normal subgroup should not change the final quotient product. :::solved-problem Let $(G,\circ)$ be a group and let $H\trianglelefteq G$. Find the identity element and inverse of $aH$ in $G/H$. ::: :::solution Let $aH\in G/H$. The identity element of $G/H$ is $$\begin{equation*} H=eH. \end{equation*}$$ The inverse of $aH$ is $$\begin{equation*} a^{-1}H. \end{equation*}$$ Indeed, $$\begin{align*} aH*a^{-1}H &=(a\circ a^{-1})H\\ &=eH\\ &=H \end{align*}$$ and $$\begin{align*} a^{-1}H*aH &=(a^{-1}\circ a)H\\ &=eH\\ &=H. \end{align*}$$ ::: We compute a product and an inverse in the finite quotient $\mathbb{Z}/n\mathbb{Z}$. Enter a modulus and one or two representatives. The calculator returns the coset product and the inverse coset. This mirrors the abstract formulas $aH*bH=(a\circ b)H$ and $(aH)^{-1}=a^{-1}H$. :::calculator[Quotient Operation Calculator] ::: :::exercise 1. Define the quotient set $G/H$. 2. State the operation used to multiply two cosets in a quotient group. 3. Explain why normality of $H$ is needed in the construction of $G/H$. 4. Find the identity element of $G/H$. 5. Find the inverse of $aH$ in $G/H$. ::: :::answer 1. $G/H=\{aH:a\in G\}$. 2. The operation is $aH*bH=(a\circ b)H$. 3. Normality ensures that the coset product does not depend on the representatives chosen. 4. The identity element is $H=eH$. 5. The inverse of $aH$ is $a^{-1}H$. ::: :::faq[Frequently Asked Questions] Q: Is $G/H$ always a group? A: No. The quotient group construction requires $H$ to be normal in $G$. Q: What is the identity element of $G/H$? A: The identity element is the coset $H=eH$. Q: Why is the quotient group also called a factor group? A: It is a group obtained by factoring out the normal subgroup $H$ from $G$. ::: :::call-to-action[Continue to Abelian and Cyclic Quotient Groups] subtitle: See how familiar group properties pass from a group to its quotient group. button: Next Lesson | published/abstract-algebra/subgroups-and-normal-subgroups/quotient-groups/abelian-and-cyclic-quotient-groups.md button-ghost: Review Normal Subgroups | published/abstract-algebra/subgroups-and-normal-subgroups/normal-subgroups/definition-and-first-examples-of-normal-subgroups.md :::