The index multiplication formula studies subgroup chains. We now study what happens when two subgroups meet. If $H$ and $K$ are subgroups of a group $G$, then $H\cap K$ is a subgroup of both $H$ and $K$. Lagrange's theorem can therefore be applied inside $H$ and inside $K$. A useful comparison is that the index of $H\cap K$ in $H$ cannot exceed the index of $K$ in $G$. This result is a precise way to measure how much of $H$ is missed by $K$. :::definition[Intersection Subgroup] Let $(G,\circ)$ be a group and let $H$ and $K$ be subgroups of $G$. The $intersection subgroup$ of $H$ and $K$ is $$\begin{equation*} H\cap K=\{x\in G:x\in H \text{ and } x\in K\}. \end{equation*}$$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ and $K$ be subgroups of $G$. Then $H\cap K$ is a subgroup of $G$. ::: :::proof Given that $(G,\circ)$ is a group and $H,K$ are subgroups of $G$. To prove that $H\cap K$ is a subgroup of $G$. Since $H$ and $K$ are subgroups of $G$, we have $e\in H$ and $e\in K$. Therefore $$\begin{equation*} e\in H\cap K. \end{equation*}$$ Let $a,b\in H\cap K$. Then $a,b\in H$ and $a,b\in K$. Since $H$ is a subgroup of $G$, $$\begin{equation*} a\circ b^{-1}\in H. \end{equation*}$$ Since $K$ is a subgroup of $G$, $$\begin{equation*} a\circ b^{-1}\in K. \end{equation*}$$ Therefore $$\begin{equation*} a\circ b^{-1}\in H\cap K. \end{equation*}$$ By the subgroup test, $H\cap K$ is a subgroup of $G$. Hence, $H\cap K$ is a subgroup of $G$. $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ and $K$ be subgroups of $G$. If $[G:K]$ is finite, then $$\begin{equation*} [H:H\cap K]\leq [G:K]. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H,K$ are subgroups of $G$, and $[G:K]$ is finite. To prove that $$\begin{equation*} [H:H\cap K]\leq [G:K]. \end{equation*}$$ Let $\mathcal{A}$ be the set of all left cosets of $H\cap K$ in $H$, and let $\mathcal{B}$ be the set of all left cosets of $K$ in $G$. Define $\Phi:\mathcal{A}\to\mathcal{B}$ by $$\begin{equation*} \Phi(h(H\cap K))=hK. \end{equation*}$$ First, we prove that $\Phi$ is well-defined. Let $$\begin{equation*} h_1(H\cap K)=h_2(H\cap K). \end{equation*}$$ Then $$\begin{equation*} h_2^{-1}\circ h_1\in H\cap K. \end{equation*}$$ Therefore $$\begin{equation*} h_2^{-1}\circ h_1\in K. \end{equation*}$$ By the left coset equality criterion, $$\begin{equation*} h_1K=h_2K. \end{equation*}$$ Thus $\Phi$ is well-defined. To prove that $\Phi$ is one-to-one, let $$\begin{equation*} \Phi(h_1(H\cap K))=\Phi(h_2(H\cap K)). \end{equation*}$$ Then $$\begin{equation*} h_1K=h_2K. \end{equation*}$$ By the left coset equality criterion, $$\begin{equation*} h_2^{-1}\circ h_1\in K. \end{equation*}$$ Since $h_1,h_2\in H$, we also have $$\begin{equation*} h_2^{-1}\circ h_1\in H. \end{equation*}$$ Therefore $$\begin{equation*} h_2^{-1}\circ h_1\in H\cap K. \end{equation*}$$ By the left coset equality criterion, $$\begin{equation*} h_1(H\cap K)=h_2(H\cap K). \end{equation*}$$ Thus $\Phi$ is one-to-one. Therefore the number of left cosets of $H\cap K$ in $H$ is less than or equal to the number of left cosets of $K$ in $G$. Hence, $$\begin{equation*} [H:H\cap K]\leq [G:K]. \end{equation*}$$ $\square$ ::: We observe the comparison between the cosets of $H\cap K$ inside $H$ and the cosets of $K$ inside $G$. Choose the two index values and watch how each coset on the left is sent into a distinct coset on the right. The map can fit only when the left side has no more cosets than the right side. Try making $[H:H\cap K]$ larger than $[G:K]$ and observe why the theorem rejects that data. :::scientific-preview[Intersection Index Injection Explorer] ::: The theorem is not only a numerical inequality; it comes from a concrete map between coset spaces. The one-to-one property prevents two different cosets of $H\cap K$ in $H$ from landing in the same coset of $K$ in $G$. :::corollary Let $(G,\circ)$ be a finite group and let $H,K$ be subgroups of $G$. Then $$\begin{equation*} \frac{|H|}{|H\cap K|}\leq \frac{|G|}{|K|}. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a finite group and $H,K$ are subgroups of $G$. To prove that $$\begin{equation*} \frac{|H|}{|H\cap K|}\leq \frac{|G|}{|K|}. \end{equation*}$$ By the previous theorem, $$\begin{equation*} [H:H\cap K]\leq [G:K]. \end{equation*}$$ By Lagrange's theorem, $$\begin{equation*} [H:H\cap K]=\frac{|H|}{|H\cap K|} \end{equation*}$$ and $$\begin{equation*} [G:K]=\frac{|G|}{|K|}. \end{equation*}$$ Therefore $$\begin{equation*} \frac{|H|}{|H\cap K|}\leq \frac{|G|}{|K|}. \end{equation*}$$ Hence, $$\begin{equation*} \frac{|H|}{|H\cap K|}\leq \frac{|G|}{|K|}. \end{equation*}$$ $\square$ ::: :::example[Intersection Index in Integers Modulo Twelve] Let $G=\mathbb{Z}_{12}$ under addition modulo $12$. Let $$\begin{equation*} H=\{0,3,6,9\} \end{equation*}$$ and $$\begin{equation*} K=\{0,4,8\}. \end{equation*}$$ Then $$\begin{equation*} H\cap K=\{0\}. \end{equation*}$$ Thus $$\begin{align*} [H:H\cap K] &=\frac{|H|}{|H\cap K|}\\ &=\frac{4}{1}\\ &=4. \end{align*}$$ Also, $$\begin{align*} [G:K] &=\frac{|G|}{|K|}\\ &=\frac{12}{3}\\ &=4. \end{align*}$$ Therefore $$\begin{equation*} [H:H\cap K]\leq [G:K]. \end{equation*}$$ ::: :::solved-problem[Bounding an Intersection Index] Let $(G,\circ)$ be a group and let $H,K$ be subgroups of $G$. Suppose $[G:K]=6$. What can be said about $[H:H\cap K]$? ::: :::solution[Solution] Let $(G,\circ)$ be a group and let $H,K$ be subgroups of $G$. Given that $$\begin{equation*} [G:K]=6. \end{equation*}$$ By the intersection index theorem, $$\begin{equation*} [H:H\cap K]\leq [G:K]. \end{equation*}$$ Therefore $$\begin{equation*} [H:H\cap K]\leq 6. \end{equation*}$$ ::: We compute the two sides of the finite version of the intersection index theorem. Enter the orders of $G$, $H$, $K$, and $H\cap K$ to calculate $[H:H\cap K]$ and $[G:K]$. The divisibility checks matter because these ratios represent subgroup indices. When all inputs are compatible, the comparison shows exactly how the intersection controls the size of $H$ relative to $K$. :::calculator[Intersection Index Calculator] ::: :::exercise[Exercises] 1. Let $G=\mathbb{Z}_{18}$, $H=\{0,3,6,9,12,15\}$, and $K=\{0,6,12\}$. Find $H\cap K$. 2. With the same $G,H,K$, compute $[H:H\cap K]$. 3. Let $H,K$ be subgroups of a finite group $G$. Prove that $|H\cap K|$ divides $|H|$ and $|K|$. ::: :::answer[Answers] 1. Since every element of $K$ belongs to $H$, we get $$\begin{equation*} H\cap K=K=\{0,6,12\}. \end{equation*}$$ 2. We have $|H|=6$ and $|H\cap K|=3$. Therefore $$\begin{equation*} [H:H\cap K]=\frac{6}{3}=2. \end{equation*}$$ 3. Since $H\cap K$ is a subgroup of $H$, Lagrange's theorem gives $$\begin{equation*} |H\cap K|\mid |H|. \end{equation*}$$ Since $H\cap K$ is a subgroup of $K$, Lagrange's theorem gives $$\begin{equation*} |H\cap K|\mid |K|. \end{equation*}$$ ::: :::faq Q: Why is $H\cap K$ a subgroup? A: It contains the identity and is closed under the subgroup test because both $H$ and $K$ are subgroups. Q: What does $[H:H\cap K]$ measure? A: It measures how many cosets of the common part $H\cap K$ are needed to cover $H$. Q: Can $[H:H\cap K]$ be larger than $[G:K]$? A: No, when $[G:K]$ is finite, there is an injection from the cosets of $H\cap K$ in $H$ into the cosets of $K$ in $G$. ::: :::call-to-action subtitle: Continue with the counting formula for products of finite subgroups. button: Next: Product Formula for Finite Subgroups | published/abstract-algebra/subgroups-and-normal-subgroups/applications-of-lagrange-theorem/product-formula-for-finite-subgroups.md :::