After studying subgroups inside quotient groups, we now count quotient groups in the finite case. Since the elements of $G/H$ are cosets of $H$, the order of $G/H$ is exactly the number of cosets of $H$ in $G$. Thus the order of a quotient group is the index of $H$ in $G$. When $G$ is finite, Lagrange's theorem converts this index into the formula $|G/H|=|G|/|H|$. This formula is simple, but students often misuse it by counting elements of $H$ instead of counting cosets. :::definition[Order of a Quotient Group] Let $(G,\circ)$ be a finite group and let $H\trianglelefteq G$. The order of the quotient group $G/H$ is the number of cosets of $H$ in $G$, denoted by $$\begin{equation*} |G/H|. \end{equation*}$$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a normal subgroup of $G$. If $G$ is finite, then $$\begin{equation*} |G/H|=\frac{|G|}{|H|}. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a normal subgroup of $G$, and $G$ is finite. To prove that $|G/H|=|G|/|H|$. Since $G/H$ is the set of all left cosets of $H$ in $G$, the number of elements of $G/H$ is the number of left cosets of $H$ in $G$. Therefore $$\begin{equation*} |G/H|=[G:H]. \end{equation*}$$ By Lagrange's theorem, $$\begin{equation*} [G:H]=\frac{|G|}{|H|}. \end{equation*}$$ Therefore $$\begin{equation*} |G/H|=\frac{|G|}{|H|}. \end{equation*}$$ Hence, $|G/H|=|G|/|H|$. $\square$ ::: :::corollary Let $(G,\circ)$ be a finite group and let $H\trianglelefteq G$. Then $$\begin{equation*} |G|=|H|\,|G/H|. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a finite group and $H\trianglelefteq G$. To prove that $|G|=|H|\,|G/H|$. By the quotient order formula, $$\begin{equation*} |G/H|=\frac{|G|}{|H|}. \end{equation*}$$ Multiplying both sides by $|H|$ gives $$\begin{equation*} |H|\,|G/H|=|G|. \end{equation*}$$ Hence, $|G|=|H|\,|G/H|$. $\square$ ::: :::example Let $G=S_3$ and let $H=A_3$. Since $$\begin{equation*} |S_3|=6 \end{equation*}$$ and $$\begin{equation*} |A_3|=3, \end{equation*}$$ we get $$\begin{equation*} |S_3/A_3|=\frac{|S_3|}{|A_3|}=\frac{6}{3}=2. \end{equation*}$$ Thus $S_3/A_3$ has two elements. Hence, $S_3/A_3$ is a quotient group of order two. $\square$ ::: :::example Let $G=\mathbb{Z}_{12}$ under addition modulo $12$, and let $$\begin{equation*} H=\{0,4,8\}. \end{equation*}$$ Then $H$ is a subgroup of the abelian group $\mathbb{Z}_{12}$, so $H\trianglelefteq \mathbb{Z}_{12}$. Since $|\mathbb{Z}_{12}|=12$ and $|H|=3$, $$\begin{equation*} |\mathbb{Z}_{12}/H|=\frac{12}{3}=4. \end{equation*}$$ Hence, the quotient group has four elements. $\square$ ::: We observe the quotient order formula by grouping elements of $G$ into equal-size cosets of $H$. Enter compatible finite orders, and the display partitions $|G|$ into blocks of size $|H|$. Each block represents one element of the quotient group $G/H$. This reinforces that $|G/H|$ counts cosets, not the elements inside one coset. :::scientific-preview[Quotient Coset Counting Explorer] ::: Enter the order of a finite group and the order of a normal subgroup. The calculator checks divisibility and returns the quotient order when the numbers are compatible with Lagrange's theorem. Try values such as $|G|=12$ and $|H|=3$, then compare with $|G|=12$ and $|H|=5$. The second input fails because a subgroup order must divide the group order in the finite case. :::calculator[Quotient Order Calculator] ::: :::solved-problem Let $G$ be a finite group of order $24$, and let $H\trianglelefteq G$ with $|H|=6$. Find $|G/H|$. ::: :::solution Let $|G|=24$ and $|H|=6$. By the quotient order formula, $$\begin{align*} |G/H| &=\frac{|G|}{|H|}\\ &=\frac{24}{6}\\ &=4. \end{align*}$$ Therefore $$\begin{equation*} |G/H|=4. \end{equation*}$$ ::: :::exercise 1. State the formula for $|G/H|$ when $G$ is finite and $H\trianglelefteq G$. 2. If $|G|=30$ and $|H|=5$, find $|G/H|$. 3. If $|G/H|=4$ and $|H|=7$, find $|G|$. 4. Find $|\mathbb{Z}_{18}/H|$ when $H$ has $6$ elements. 5. Explain why $|G|=20$ and $|H|=6$ cannot occur for a finite subgroup $H$ of $G$. ::: :::answer 1. $|G/H|=|G|/|H|$. 2. $|G/H|=30/5=6$. 3. $|G|=|H|\,|G/H|=7\cdot 4=28$. 4. $|\mathbb{Z}_{18}/H|=18/6=3$. 5. By Lagrange's theorem, $|H|$ must divide $|G|$, but $6$ does not divide $20$. ::: :::faq[Frequently Asked Questions] Q: Why is $|G/H|$ equal to an index? A: The elements of $G/H$ are precisely the cosets of $H$ in $G$. Q: Does the quotient order formula require $G$ to be finite? A: The formula $|G/H|=|G|/|H|$ is a finite-group formula. For infinite groups, index language is safer. Q: What is counted by $|G/H|$? A: It counts cosets of $H$ in $G$, not elements inside one coset. ::: :::call-to-action[Continue to Abelian Quotients and Commutators] subtitle: Connect abelian quotient groups with commutator elements in the normal subgroup. button: Next Lesson | published/abstract-algebra/subgroups-and-normal-subgroups/quotient-groups/abelian-quotients-and-commutators.md button-ghost: Previous Lesson | published/abstract-algebra/subgroups-and-normal-subgroups/quotient-groups/normal-subgroups-in-quotient-groups.md :::