We now study the most important structural fact about cosets. Two cosets of the same subgroup cannot partly overlap. If they share even one element, then they are exactly the same coset. If they do not share any element, then they are disjoint. This is the reason cosets behave like separate blocks inside a group. Students often expect two translated copies of a subgroup to overlap partially, but group inverses prevent that from happening. :::lemma Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $a\in G-H$, then $$\begin{equation*} aH\cap H=\varnothing. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $a\in G-H$. To prove that $aH\cap H=\varnothing$. If possible let $$\begin{equation*} aH\cap H\neq\varnothing. \end{equation*}$$ Then there exists $x\in G$ such that $$\begin{equation*} x\in aH\cap H. \end{equation*}$$ Therefore $x\in aH$ and $x\in H$. Since $x\in aH$, there exists $h\in H$ such that $$\begin{equation*} x=a\circ h. \end{equation*}$$ Then $$\begin{align*} x=a\circ h &\implies x\circ h^{-1}=a\circ h\circ h^{-1}\\ &\implies x\circ h^{-1}=a\circ e\\ &\implies x\circ h^{-1}=a. \end{align*}$$ Since $x\in H$ and $h^{-1}\in H$, we get $x\circ h^{-1}\in H$. Therefore $a\in H$. This contradicts $a\in G-H$. Hence, $$\begin{equation*} aH\cap H=\varnothing. \end{equation*}$$ $\square$ ::: :::lemma Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $a\in G-H$, then $$\begin{equation*} Ha\cap H=\varnothing. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $a\in G-H$. To prove that $Ha\cap H=\varnothing$. If possible let $$\begin{equation*} Ha\cap H\neq\varnothing. \end{equation*}$$ Then there exists $x\in G$ such that $$\begin{equation*} x\in Ha\cap H. \end{equation*}$$ Therefore $x\in Ha$ and $x\in H$. Since $x\in Ha$, there exists $h\in H$ such that $$\begin{equation*} x=h\circ a. \end{equation*}$$ Then $$\begin{align*} x=h\circ a &\implies h^{-1}\circ x=h^{-1}\circ h\circ a\\ &\implies h^{-1}\circ x=e\circ a\\ &\implies h^{-1}\circ x=a. \end{align*}$$ Since $h^{-1}\in H$ and $x\in H$, we get $h^{-1}\circ x\in H$. Therefore $a\in H$. This contradicts $a\in G-H$. Hence, $$\begin{equation*} Ha\cap H=\varnothing. \end{equation*}$$ $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $a,b\in G$, then either $$\begin{equation*} aH=bH \end{equation*}$$ or $$\begin{equation*} aH\cap bH=\varnothing. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $a,b\in G$. To prove that either $aH=bH$ or $aH\cap bH=\varnothing$. Assume that $$\begin{equation*} aH\cap bH\neq\varnothing. \end{equation*}$$ To prove that $aH=bH$. Let $x\in aH\cap bH$. Then there exist $h_1,h_2\in H$ such that $$\begin{equation*} x=a\circ h_1=b\circ h_2. \end{equation*}$$ Therefore $$\begin{align*} a\circ h_1=b\circ h_2 &\implies a\circ h_1\circ h_1^{-1}=b\circ h_2\circ h_1^{-1}\\ &\implies a=b\circ (h_2\circ h_1^{-1}). \end{align*}$$ Let $y\in aH$. Then there exists $h\in H$ such that $$\begin{equation*} y=a\circ h. \end{equation*}$$ Using $a=b\circ (h_2\circ h_1^{-1})$, we get $$\begin{align*} y &=a\circ h\\ &=b\circ (h_2\circ h_1^{-1})\circ h\\ &=b\circ (h_2\circ h_1^{-1}\circ h). \end{align*}$$ Since $h_2,h_1^{-1},h\in H$, we get $h_2\circ h_1^{-1}\circ h\in H$. Therefore $y\in bH$. Therefore $$\begin{equation*} aH\subseteq bH. \end{equation*}$$ Similarly, by interchanging $a$ and $b$, we get $$\begin{equation*} bH\subseteq aH. \end{equation*}$$ Therefore $aH=bH$. Hence, either $aH=bH$ or $aH\cap bH=\varnothing$. $\square$ ::: We observe the equal-or-disjoint theorem by comparing two cosets in $\mathbb{Z}_n$. Choose two representatives and watch the intersection of the two cosets. If the intersection has even one element, the two cosets become exactly equal. If the cosets are not equal, their intersection is empty, so partial overlap never occurs. :::scientific-preview[Equal or Disjoint Cosets Explorer] ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $a,b\in G$, then either $$\begin{equation*} Ha=Hb \end{equation*}$$ or $$\begin{equation*} Ha\cap Hb=\varnothing. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $a,b\in G$. To prove that either $Ha=Hb$ or $Ha\cap Hb=\varnothing$. Assume that $$\begin{equation*} Ha\cap Hb\neq\varnothing. \end{equation*}$$ To prove that $Ha=Hb$. Let $x\in Ha\cap Hb$. Then there exist $h_1,h_2\in H$ such that $$\begin{equation*} x=h_1\circ a=h_2\circ b. \end{equation*}$$ Therefore $$\begin{align*} h_1\circ a=h_2\circ b &\implies h_1^{-1}\circ h_1\circ a=h_1^{-1}\circ h_2\circ b\\ &\implies a=(h_1^{-1}\circ h_2)\circ b. \end{align*}$$ Let $y\in Ha$. Then there exists $h\in H$ such that $$\begin{equation*} y=h\circ a. \end{equation*}$$ Using $a=(h_1^{-1}\circ h_2)\circ b$, we get $$\begin{align*} y &=h\circ a\\ &=h\circ (h_1^{-1}\circ h_2)\circ b\\ &=(h\circ h_1^{-1}\circ h_2)\circ b. \end{align*}$$ Since $h,h_1^{-1},h_2\in H$, we get $h\circ h_1^{-1}\circ h_2\in H$. Therefore $y\in Hb$. Therefore $$\begin{equation*} Ha\subseteq Hb. \end{equation*}$$ Similarly, by interchanging $a$ and $b$, we get $$\begin{equation*} Hb\subseteq Ha. \end{equation*}$$ Therefore $Ha=Hb$. Hence, either $Ha=Hb$ or $Ha\cap Hb=\varnothing$. $\square$ ::: :::example[Disjoint Cosets in Integers Modulo Twelve] Let $G=\mathbb{Z}_{12}$ under addition modulo $12$ and let $$\begin{equation*} H=\{0,4,8\}. \end{equation*}$$ The distinct cosets of $H$ are $$\begin{align*} 0+H&=\{0,4,8\},\\ 1+H&=\{1,5,9\},\\ 2+H&=\{2,6,10\},\\ 3+H&=\{3,7,11\}. \end{align*}$$ Any two of these cosets are disjoint. For example, $$\begin{equation*} (1+H)\cap (2+H)=\varnothing. \end{equation*}$$ Also, $$\begin{equation*} 4+H=H, \end{equation*}$$ because $4\in H$. ::: :::solved-problem[Using One Common Element] Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Suppose $aH\cap bH\neq\varnothing$. Prove that $b\in aH$. ::: :::solution[Solution] Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Given that $aH\cap bH\neq\varnothing$. By the equal or disjoint theorem for left cosets, we get $$\begin{equation*} aH=bH. \end{equation*}$$ Since $e\in H$, we get $$\begin{align*} b &=b\circ e\\ &\in bH\\ &=aH. \end{align*}$$ Therefore $b\in aH$. ::: We compute two modular cosets and their intersection directly. This lets us test the equal-or-disjoint theorem without relying on a diagram. If the intersection is nonempty, the calculator reports equality; otherwise it reports disjointness. The result should never be a nonempty proper overlap. :::calculator[Coset Intersection Tester] ::: :::exercise[Exercises] 1. Let $G=\mathbb{Z}_{9}$ under addition modulo $9$ and let $H=\{0,3,6\}$. List all distinct cosets of $H$. 2. Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $aH\cap H\neq\varnothing$, prove that $a\in H$. 3. Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $Ha\cap Hb\neq\varnothing$, prove that $Ha=Hb$. ::: :::answer[Answers] 1. The distinct cosets are $$\begin{align*} 0+H&=\{0,3,6\},\\ 1+H&=\{1,4,7\},\\ 2+H&=\{2,5,8\}. \end{align*}$$ 2. Since $aH\cap H\neq\varnothing$ and $H=eH$, the equal or disjoint theorem gives $aH=H$. Hence, $a\in H$. 3. This is exactly the right coset equal or disjoint theorem. Since the intersection is nonempty, the disjoint alternative is impossible. Therefore $Ha=Hb$. ::: :::faq Q: Can two cosets overlap at one element only? A: No. If two cosets of the same subgroup have one common element, then they are equal. Q: Why is this theorem important? A: It shows that cosets divide the group into non-overlapping blocks. Q: Does the same equal-or-disjoint result hold for right cosets? A: Yes. The proof is similar, but multiplication must be placed on the correct side. ::: :::call-to-action subtitle: Continue with the partition of a group by its left and right cosets. button: Next: Cosets as Partitions | published/abstract-algebra/introduction-to-groups/cosets-of-subgroups/cosets-as-partitions.md :::