After proving Lagrange's theorem, we now use it to compare several subgroups lying one inside another. If $K$ is a subgroup of $H$ and $H$ is a subgroup of $G$, then the index from $K$ to $G$ can be computed in two stages. First count how many cosets of $H$ are needed in $G$, and then count how many cosets of $K$ are needed in $H$. The product gives the number of cosets of $K$ in $G$. This is the index multiplication formula, and it is one of the most useful counting tools for subgroup chains. :::definition[Subgroup Chain] Let $(G,\circ)$ be a group. A $subgroup chain$ in $G$ is a sequence of subgroups $$\begin{equation*} K\leq H\leq G \end{equation*}$$ where $K$ is a subgroup of $H$ and $H$ is a subgroup of $G$. ::: :::theorem[Index Multiplication Formula] Let $(G,\circ)$ be a finite group. If $K$ and $H$ are subgroups of $G$ such that $$\begin{equation*} K\leq H\leq G, \end{equation*}$$ then $$\begin{equation*} [G:K]=[G:H][H:K]. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a finite group and $K,H$ are subgroups of $G$ such that $K\leq H\leq G$. To prove that $$\begin{equation*} [G:K]=[G:H][H:K]. \end{equation*}$$ By Lagrange's theorem applied to $K\leq G$, we get $$\begin{equation*} [G:K]=\frac{|G|}{|K|}. \end{equation*}$$ By Lagrange's theorem applied to $H\leq G$, we get $$\begin{equation*} [G:H]=\frac{|G|}{|H|}. \end{equation*}$$ By Lagrange's theorem applied to $K\leq H$, we get $$\begin{equation*} [H:K]=\frac{|H|}{|K|}. \end{equation*}$$ Therefore $$\begin{align*} [G:H][H:K] &=\frac{|G|}{|H|}\cdot \frac{|H|}{|K|}\\ &=\frac{|G|}{|K|}\\ &=[G:K]. \end{align*}$$ Hence, $$\begin{equation*} [G:K]=[G:H][H:K]. \end{equation*}$$ $\square$ ::: We observe the subgroup chain $K\leq H\leq G$ by separating the count into two stages. Choose compatible orders for $G$, $H$, and $K$, and notice how the index from $G$ down to $K$ factors through the middle subgroup $H$. As $|H|$ changes, the two smaller indices change in opposite ways, but their product remains the full index $[G:K]$. Try setting $|G|=12$, $|H|=4$, and $|K|=2$ and observe the formula from the example. :::scientific-preview[Subgroup Chain Index Explorer] ::: The visual count separates one large quotient into two smaller quotient counts. This is exactly the cancellation pattern in the proof, but now it is tied to the number of cosets at each level of the chain. :::corollary Let $(G,\circ)$ be a finite group and let $K,H$ be subgroups of $G$ such that $K\leq H\leq G$. If $[G:K]$ is prime, then either $$\begin{equation*} H=K \end{equation*}$$ or $$\begin{equation*} H=G. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a finite group, $K,H$ are subgroups of $G$ such that $K\leq H\leq G$, and $[G:K]$ is prime. To prove that either $H=K$ or $H=G$. Let $$\begin{equation*} [G:K]=p, \end{equation*}$$ where $p$ is prime. By the index multiplication formula, $$\begin{equation*} [G:K]=[G:H][H:K]. \end{equation*}$$ Therefore $$\begin{equation*} p=[G:H][H:K]. \end{equation*}$$ Since $p$ is prime and $[G:H]$ and $[H:K]$ are positive integers, we get either $$\begin{equation*} [G:H]=1 \end{equation*}$$ or $$\begin{equation*} [H:K]=1. \end{equation*}$$ If $[G:H]=1$, then $H=G$. If $[H:K]=1$, then $H=K$. Hence, either $H=K$ or $H=G$. $\square$ ::: :::example[A Chain of Subgroups in Integers Modulo Twelve] Let $G=\mathbb{Z}_{12}$ under addition modulo $12$. Let $$\begin{equation*} H=\{0,3,6,9\} \end{equation*}$$ and $$\begin{equation*} K=\{0,6\}. \end{equation*}$$ Then $$\begin{equation*} K\leq H\leq G. \end{equation*}$$ Now $$\begin{align*} [G:H]&=\frac{12}{4}=3,\\ [H:K]&=\frac{4}{2}=2,\\ [G:K]&=\frac{12}{2}=6. \end{align*}$$ Therefore $$\begin{equation*} [G:K]=[G:H][H:K]. \end{equation*}$$ ::: :::solved-problem[Finding an Unknown Index] Let $(G,\circ)$ be a finite group and let $K\leq H\leq G$. Suppose $$\begin{equation*} [G:K]=30 \end{equation*}$$ and $$\begin{equation*} [G:H]=5. \end{equation*}$$ Find $[H:K]$. ::: :::solution[Solution] Let $(G,\circ)$ be a finite group and let $K\leq H\leq G$. By the index multiplication formula, $$\begin{equation*} [G:K]=[G:H][H:K]. \end{equation*}$$ Substituting the given values, we get $$\begin{align*} 30&=5[H:K]\\ &\implies [H:K]=6. \end{align*}$$ Therefore $$\begin{equation*} [H:K]=6. \end{equation*}$$ ::: We use the index multiplication formula as a numerical tool when two of the three indices are known. Enter any two values and leave one value blank to solve for the missing index. Notice that the unknown must be a positive integer, because every index counts cosets. This helps distinguish valid subgroup-chain data from numbers that cannot arise from the formula. :::calculator[Index Multiplication Calculator] ::: :::exercise[Exercises] 1. Let $K\leq H\leq G$ and suppose $[G:H]=4$ and $[H:K]=3$. Find $[G:K]$. 2. Let $K\leq H\leq G$ and suppose $[G:K]=24$ and $[H:K]=6$. Find $[G:H]$. 3. Let $K\leq H\leq G$ and suppose $[G:K]=7$. Prove that $H=K$ or $H=G$. ::: :::answer[Answers] 1. By the index multiplication formula, $$\begin{equation*} [G:K]=[G:H][H:K]=4\cdot 3=12. \end{equation*}$$ 2. Since $$\begin{equation*} [G:K]=[G:H][H:K], \end{equation*}$$ we get $$\begin{align*} 24&=[G:H]\cdot 6\\ &\implies [G:H]=4. \end{align*}$$ 3. Since $7$ is prime and $$\begin{equation*} 7=[G:H][H:K], \end{equation*}$$ one of the two positive factors must be $1$. Hence, $H=G$ or $H=K$. ::: :::faq Q: What does the index multiplication formula count? A: It counts cosets through a subgroup chain $K\leq H\leq G$. Q: Why does the formula resemble cancellation of fractions? A: In finite groups, the formula follows from $[G:H]=|G|/|H|$ and $[H:K]=|H|/|K|$. Q: What happens if $[G:K]$ is prime? A: There is no proper intermediate subgroup strictly between $K$ and $G$. ::: :::call-to-action subtitle: Continue with the way subgroup intersections control index. button: Next: Index and Intersections of Subgroups | published/abstract-algebra/subgroups-and-normal-subgroups/applications-of-lagrange-theorem/index-and-intersections-of-subgroups.md :::