We finish this section by connecting cosets with intersections of subgroups. If $H$ and $K$ are subgroups of the same group, then $H\cap K$ is also a subgroup. Its cosets can be described as intersections of corresponding cosets of $H$ and $K$. This result is useful because it lets us move between subgroup intersections and coset intersections without guessing element by element. The formulas in this lecture also train students to keep left and right multiplication separate. :::theorem Let $(G,\circ)$ be a group and let $H$ and $K$ be subgroups of $G$. If $a\in G$, then $$\begin{equation*} (H\cap K)a=Ha\cap Ka. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ and $K$ are subgroups of $G$, and $a\in G$. To prove that $(H\cap K)a=Ha\cap Ka$. [1] To prove that $(H\cap K)a\subseteq Ha\cap Ka$. Let $x\in (H\cap K)a$. Then there exists $u\in H\cap K$ such that $$\begin{equation*} x=u\circ a. \end{equation*}$$ Since $u\in H\cap K$, we get $u\in H$ and $u\in K$. Since $u\in H$, we get $x=u\circ a\in Ha$. Since $u\in K$, we get $x=u\circ a\in Ka$. Therefore $x\in Ha\cap Ka$. Thus $$\begin{equation*} (H\cap K)a\subseteq Ha\cap Ka. \end{equation*}$$ [2] To prove that $Ha\cap Ka\subseteq (H\cap K)a$. Let $x\in Ha\cap Ka$. Then $x\in Ha$ and $x\in Ka$. Since $x\in Ha$, there exists $h\in H$ such that $$\begin{equation*} x=h\circ a. \end{equation*}$$ Since $x\in Ka$, there exists $k\in K$ such that $$\begin{equation*} x=k\circ a. \end{equation*}$$ Therefore $$\begin{align*} h\circ a=k\circ a &\implies h\circ a\circ a^{-1}=k\circ a\circ a^{-1}\\ &\implies h=k. \end{align*}$$ Since $h=k$, $h\in H$, and $k\in K$, we get $h\in H\cap K$. Therefore $$\begin{equation*} x=h\circ a\in (H\cap K)a. \end{equation*}$$ Thus $$\begin{equation*} Ha\cap Ka\subseteq (H\cap K)a. \end{equation*}$$ Hence, $$\begin{equation*} (H\cap K)a=Ha\cap Ka. \end{equation*}$$ $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ and $K$ be subgroups of $G$. If $a\in G$, then $$\begin{equation*} a(H\cap K)=aH\cap aK. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ and $K$ are subgroups of $G$, and $a\in G$. To prove that $a(H\cap K)=aH\cap aK$. [1] To prove that $a(H\cap K)\subseteq aH\cap aK$. Let $x\in a(H\cap K)$. Then there exists $u\in H\cap K$ such that $$\begin{equation*} x=a\circ u. \end{equation*}$$ Since $u\in H\cap K$, we get $u\in H$ and $u\in K$. Since $u\in H$, we get $x=a\circ u\in aH$. Since $u\in K$, we get $x=a\circ u\in aK$. Therefore $x\in aH\cap aK$. Thus $$\begin{equation*} a(H\cap K)\subseteq aH\cap aK. \end{equation*}$$ [2] To prove that $aH\cap aK\subseteq a(H\cap K)$. Let $x\in aH\cap aK$. Then $x\in aH$ and $x\in aK$. Since $x\in aH$, there exists $h\in H$ such that $$\begin{equation*} x=a\circ h. \end{equation*}$$ Since $x\in aK$, there exists $k\in K$ such that $$\begin{equation*} x=a\circ k. \end{equation*}$$ Therefore $$\begin{align*} a\circ h=a\circ k &\implies a^{-1}\circ a\circ h=a^{-1}\circ a\circ k\\ &\implies h=k. \end{align*}$$ Since $h=k$, $h\in H$, and $k\in K$, we get $h\in H\cap K$. Therefore $$\begin{equation*} x=a\circ h\in a(H\cap K). \end{equation*}$$ Thus $$\begin{equation*} aH\cap aK\subseteq a(H\cap K). \end{equation*}$$ Hence, $$\begin{equation*} a(H\cap K)=aH\cap aK. \end{equation*}$$ $\square$ ::: We observe the formula $a(H\cap K)=aH\cap aK$ in a modular group. Choose two subgroups and a representative, then compare the translated intersection with the intersection of the translated cosets. The two sets match because the same element must come from both subgroup translations. This helps connect subgroup intersections with coset intersections without guessing element by element. :::scientific-preview[Coset Intersection Formula Explorer] ::: :::example[Intersections of Cosets in Integers Modulo Twelve] Let $G=\mathbb{Z}_{12}$ under addition modulo $12$. Let $$\begin{equation*} H=\{0,4,8\} \end{equation*}$$ and $$\begin{equation*} K=\{0,6\}. \end{equation*}$$ Then $$\begin{equation*} H\cap K=\{0\}. \end{equation*}$$ For $a=2$, we have $$\begin{align*} 2+(H\cap K) &=\{2\}. \end{align*}$$ Also, $$\begin{align*} 2+H&=\{2,6,10\},\\ 2+K&=\{2,8\}. \end{align*}$$ Therefore $$\begin{align*} (2+H)\cap (2+K) &=\{2\}\\ &=2+(H\cap K). \end{align*}$$ ::: :::solved-problem[A Right Coset Intersection] Let $(G,\circ)$ be a group and let $H,K$ be subgroups of $G$. Suppose $x\in Ha\cap Ka$. Prove that $x\in (H\cap K)a$. ::: :::solution[Solution] Let $(G,\circ)$ be a group, let $H,K$ be subgroups of $G$, and let $a\in G$. Given that $x\in Ha\cap Ka$. Then $x\in Ha$ and $x\in Ka$. Hence, there exist $h\in H$ and $k\in K$ such that $$\begin{equation*} x=h\circ a=k\circ a. \end{equation*}$$ Therefore $$\begin{align*} h\circ a=k\circ a &\implies h\circ a\circ a^{-1}=k\circ a\circ a^{-1}\\ &\implies h=k. \end{align*}$$ Since $h\in H$ and $h=k\in K$, we get $h\in H\cap K$. Therefore $$\begin{equation*} x=h\circ a\in (H\cap K)a. \end{equation*}$$ ::: We compute both sides of the modular intersection formula. Enter subgroup steps for $H$ and $K$ and a representative $a$, then compare $a+(H\cap K)$ with $(a+H)\cap(a+K)$. The output also displays $H\cap K$ so the common subgroup is visible before translation. This makes the formula a direct set calculation. :::calculator[Subgroup Coset Intersection Calculator] ::: :::exercise[Exercises] 1. Let $G=\mathbb{Z}_{12}$, $H=\{0,3,6,9\}$, and $K=\{0,6\}$. Find $2+(H\cap K)$. 2. With the same $G,H,K$, find $(2+H)\cap (2+K)$. 3. Let $(G,\circ)$ be a group and let $H,K$ be subgroups of $G$. Prove that $a(H\cap K)\subseteq aH\cap aK$. ::: :::answer[Answers] 1. Since $$\begin{equation*} H\cap K=\{0,6\}, \end{equation*}$$ we get $$\begin{equation*} 2+(H\cap K)=\{2,8\}. \end{equation*}$$ 2. We have $$\begin{align*} 2+H&=\{2,5,8,11\},\\ 2+K&=\{2,8\}. \end{align*}$$ Therefore $$\begin{equation*} (2+H)\cap (2+K)=\{2,8\}. \end{equation*}$$ 3. If $x\in a(H\cap K)$, then $x=a\circ u$ for some $u\in H\cap K$. Hence $u\in H$ and $u\in K$. Therefore $x\in aH$ and $x\in aK$, so $x\in aH\cap aK$. ::: :::faq Q: Why does $(H\cap K)a$ equal $Ha\cap Ka$? A: An element is in both $Ha$ and $Ka$ exactly when the multiplying element before $a$ belongs to both $H$ and $K$. Q: Does a similar formula hold for left cosets? A: Yes. We have $a(H\cap K)=aH\cap aK$. Q: Is $H\cap K$ always a subgroup? A: Yes. The intersection of two subgroups of the same group is again a subgroup. ::: :::call-to-action subtitle: Continue with index and the counting theorem that grows out of coset partitions. button: Next Section: Index and Lagrange Theorem | published/abstract-algebra/introduction-to-groups/index-and-lagrange-theorem/index-of-a-subgroup.md :::