The index of a subgroup counts how many cosets the subgroup produces. We now combine this with the fact that all cosets of a subgroup have the same number of elements. This gives one of the central counting results in finite group theory: Lagrange's theorem. The theorem says that the order of a subgroup must divide the order of the finite group. Many later restrictions on groups, elements, and subgroup structures come from this single theorem. :::definition[Order of a Finite Group] Let $(G,\circ)$ be a finite group. The $order$ of $G$ is the number of elements in $G$ and is denoted by $$\begin{equation*} |G|. \end{equation*}$$ ::: :::theorem[Lagrange's Theorem] Let $(G,\circ)$ be a finite group and let $H$ be a subgroup of $G$. Then $|H|$ divides $|G|$ and $$\begin{equation*} [G:H]=\frac{|G|}{|H|}. \end{equation*}$$ Equivalently, $$\begin{equation*} |G|=[G:H]|H|. \end{equation*}$$ ::: Lagrange's theorem says that a finite group is divided into equal-sized coset blocks. The size of each block is $|H|$, and the number of blocks is $[G:H]$. Change the subgroup size and the index to see how their product gives the whole group order. The picture makes the formula $|G|=[G:H]|H|$ behave like ordinary multiplication of equal blocks. :::scientific-preview[Lagrange Block Counter] ::: The cosets fill the group without overlap and without gaps. That is why the total number of elements must be a multiple of the subgroup order. :::proof Given that $(G,\circ)$ is a finite group and $H$ is a subgroup of $G$. To prove that $|H|$ divides $|G|$ and $$\begin{equation*} [G:H]=\frac{|G|}{|H|}. \end{equation*}$$ Let the distinct left cosets of $H$ in $G$ be $$\begin{equation*} a_1H,a_2H,\ldots,a_rH. \end{equation*}$$ Then $$\begin{equation*} [G:H]=r. \end{equation*}$$ Since the left cosets of $H$ form a partition of $G$, we get $$\begin{equation*} G=a_1H\cup a_2H\cup\cdots\cup a_rH, \end{equation*}$$ where the cosets $a_1H,a_2H,\ldots,a_rH$ are pairwise disjoint. Since every left coset of $H$ has the same cardinality as $H$, we get $$\begin{equation*} |a_iH|=|H| \quad \text{for } i=1,2,\ldots,r. \end{equation*}$$ Therefore $$\begin{align*} |G| &=|a_1H|+|a_2H|+\cdots+|a_rH|\\ &=|H|+|H|+\cdots+|H|\\ &=r|H|\\ &=[G:H]|H|. \end{align*}$$ Therefore $|H|$ divides $|G|$. Also, $$\begin{align*} |G|=[G:H]|H| &\implies [G:H]=\frac{|G|}{|H|}. \end{align*}$$ Hence, $|H|$ divides $|G|$ and $[G:H]=\frac{|G|}{|H|}$. $\square$ ::: Lagrange's theorem is a divisibility theorem, not an existence theorem. It says that if $H$ is a subgroup of $G$, then $|H|$ must be a divisor of $|G|$. It does not say that every divisor of $|G|$ is the order of a subgroup. This distinction is important and prevents a common error in first courses on group theory. The formula in Lagrange's theorem has three quantities: $|G|$, $|H|$, and $[G:H]$. Enter any two of them and leave the third blank to solve for the missing value. The solver checks divisibility when the whole group order is known. Use it to test whether proposed subgroup data can be consistent with Lagrange's theorem. :::calculator[Lagrange Formula Solver] ::: :::corollary Let $(G,\circ)$ be a finite group and let $H$ be a subgroup of $G$. Then $$\begin{equation*} |G|=[G:H]|H|. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a finite group and $H$ is a subgroup of $G$. To prove that $|G|=[G:H]|H|$. By Lagrange's theorem, $$\begin{equation*} [G:H]=\frac{|G|}{|H|}. \end{equation*}$$ Therefore $$\begin{align*} [G:H]=\frac{|G|}{|H|} &\implies |G|=[G:H]|H|. \end{align*}$$ Hence, $$\begin{equation*} |G|=[G:H]|H|. \end{equation*}$$ $\square$ ::: :::example[Lagrange's Theorem in Integers Modulo Twelve] Let $G=\mathbb{Z}_{12}$ and let $$\begin{equation*} H=\{0,4,8\}. \end{equation*}$$ Then $|G|=12$ and $|H|=3$. By Lagrange's theorem, $$\begin{align*} [G:H] &=\frac{|G|}{|H|}\\ &=\frac{12}{3}\\ &=4. \end{align*}$$ This agrees with the four cosets $$\begin{equation*} H,\quad 1+H,\quad 2+H,\quad 3+H. \end{equation*}$$ ::: :::solved-problem[Finding the Index from Orders] Let $(G,\circ)$ be a finite group with $|G|=60$. Let $H$ be a subgroup of $G$ such that $|H|=12$. Find $[G:H]$. ::: :::solution[Solution] Let $(G,\circ)$ be a finite group with $|G|=60$ and let $H$ be a subgroup of $G$ such that $|H|=12$. By Lagrange's theorem, $$\begin{align*} [G:H] &=\frac{|G|}{|H|}\\ &=\frac{60}{12}\\ &=5. \end{align*}$$ Therefore $$\begin{equation*} [G:H]=5. \end{equation*}$$ ::: :::solved-problem[Finding the Order of a Subgroup] Let $(G,\circ)$ be a finite group with $|G|=48$. Suppose $H$ is a subgroup of $G$ and $[G:H]=6$. Find $|H|$. ::: :::solution[Solution] Let $(G,\circ)$ be a finite group with $|G|=48$. Let $H$ be a subgroup of $G$ and suppose $[G:H]=6$. By Lagrange's theorem, $$\begin{equation*} |G|=[G:H]|H|. \end{equation*}$$ Therefore $$\begin{align*} 48&=6|H|\\ &\implies |H|=8. \end{align*}$$ Hence, $$\begin{equation*} |H|=8. \end{equation*}$$ ::: :::exercise[Exercises] 1. Let $|G|=72$ and let $H$ be a subgroup of $G$ with $|H|=9$. Find $[G:H]$. 2. Let $|G|=84$ and let $[G:H]=7$. Find $|H|$. 3. Let $H$ be a subgroup of a finite group $G$. If $|G|=45$ and $|H|=10$, explain why this is impossible. ::: :::answer[Answers] 1. By Lagrange's theorem, $$\begin{equation*} [G:H]=\frac{72}{9}=8. \end{equation*}$$ 2. Since $|G|=[G:H]|H|$, we get $$\begin{equation*} 84=7|H|. \end{equation*}$$ Therefore $$\begin{equation*} |H|=12. \end{equation*}$$ 3. If $H$ is a subgroup of $G$, then $|H|$ must divide $|G|$. But $10$ does not divide $45$. Therefore such a subgroup cannot exist. ::: :::faq Q: What does Lagrange's theorem say? A: It says that the order of a subgroup divides the order of a finite group. Q: Does every divisor of $|G|$ occur as the order of a subgroup? A: No. Lagrange's theorem gives a necessary condition, not a sufficient condition. Q: Why do cosets appear in the proof? A: Cosets partition the group into equal-sized blocks, and each block has $|H|$ elements. ::: :::call-to-action subtitle: Continue with the direct restrictions Lagrange's theorem places on possible subgroup orders. button: Next: Divisors and Subgroup Orders | published/abstract-algebra/introduction-to-groups/index-and-lagrange-theorem/divisors-and-subgroup-orders.md :::