After studying applications of Lagrange's theorem, we now return to cosets with a wider construction. A left coset multiplies a subgroup on the left of an element, and a right coset multiplies a subgroup on the right of an element. A double coset allows both actions at the same time. If $H$ and $K$ are subgroups of a group $G$, then the double coset $HaK$ consists of all elements obtained by multiplying $a$ on the left by elements of $H$ and on the right by elements of $K$. In this lecture, we define double cosets, compare them with ordinary cosets, and compute examples that show why the construction is useful. :::definition[Double Coset] Let $(G,\circ)$ be a group and let $H$ and $K$ be subgroups of $G$. For $a\in G$, the $double coset$ of $a$ with respect to $H$ and $K$ is the subset $$\begin{equation*} HaK=\{h\circ a\circ k:h\in H,\;k\in K\}. \end{equation*}$$ The element $a$ is called a representative of the double coset $HaK$. ::: The notation $HaK$ should be read carefully. It is not the same as $aHK$, and it is not usually the same as $HKa$. The order of multiplication matters in a nonabelian group. Even in an abelian group, the double coset is conceptually different because two subgroups are involved. A common mistake is to treat $HaK$ as a product of three subgroups. This is not correct, because $a$ is an element, not necessarily a subgroup. :::theorem Let $(G,\circ)$ be a group and let $H$ and $K$ be subgroups of $G$. If $a\in G$, then $$\begin{equation*} a\in HaK. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ and $K$ are subgroups of $G$, and $a\in G$. To prove that $$\begin{equation*} a\in HaK. \end{equation*}$$ Since $H$ and $K$ are subgroups of $G$, the identity element $e$ of $G$ belongs to both $H$ and $K$. Now $$\begin{align*} a &=e\circ a\circ e &&[\because e \text{ is the identity element}]\\ &\in HaK &&[\because e\in H \text{ and } e\in K]. \end{align*}$$ Hence, $$\begin{equation*} a\in HaK. \end{equation*}$$ $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ and $K$ be subgroups of $G$. If $a\in G$, then $$\begin{equation*} HaK=\bigcup_{h\in H}haK \end{equation*}$$ and $$\begin{equation*} HaK=\bigcup_{k\in K}Hak. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ and $K$ are subgroups of $G$, and $a\in G$. To prove that $$\begin{equation*} HaK=\bigcup_{h\in H}haK \end{equation*}$$ and $$\begin{equation*} HaK=\bigcup_{k\in K}Hak. \end{equation*}$$ [1] To prove that $HaK=\bigcup_{h\in H}haK$. Let $x\in HaK$. Then there exist $h\in H$ and $k\in K$ such that $$\begin{equation*} x=h\circ a\circ k. \end{equation*}$$ Therefore $x\in haK$ for some $h\in H$. Hence, $$\begin{equation*} HaK\subseteq \bigcup_{h\in H}haK. \end{equation*}$$ Let $x\in \bigcup_{h\in H}haK$. Then there exists $h\in H$ such that $$\begin{equation*} x\in haK. \end{equation*}$$ Therefore there exists $k\in K$ such that $$\begin{equation*} x=h\circ a\circ k. \end{equation*}$$ Thus $x\in HaK$. Therefore $$\begin{equation*} \bigcup_{h\in H}haK\subseteq HaK. \end{equation*}$$ Therefore $$\begin{equation*} HaK=\bigcup_{h\in H}haK. \end{equation*}$$ [2] To prove that $HaK=\bigcup_{k\in K}Hak$. Let $x\in HaK$. Then there exist $h\in H$ and $k\in K$ such that $$\begin{equation*} x=h\circ a\circ k. \end{equation*}$$ Therefore $x\in Hak$ for some $k\in K$. Hence, $$\begin{equation*} HaK\subseteq \bigcup_{k\in K}Hak. \end{equation*}$$ Let $x\in \bigcup_{k\in K}Hak$. Then there exists $k\in K$ such that $$\begin{equation*} x\in Hak. \end{equation*}$$ Therefore there exists $h\in H$ such that $$\begin{equation*} x=h\circ a\circ k. \end{equation*}$$ Thus $x\in HaK$. Therefore $$\begin{equation*} \bigcup_{k\in K}Hak\subseteq HaK. \end{equation*}$$ Hence, $$\begin{equation*} HaK=\bigcup_{h\in H}haK \end{equation*}$$ and $$\begin{equation*} HaK=\bigcup_{k\in K}Hak. \end{equation*}$$ $\square$ ::: :::corollary Let $(G,\circ)$ be a group and let $H$ and $K$ be subgroups of $G$. If $a\in G$, then $$\begin{equation*} HaK=H(aK) \end{equation*}$$ and $$\begin{equation*} HaK=(Ha)K. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ and $K$ are subgroups of $G$, and $a\in G$. To prove that $$\begin{equation*} HaK=H(aK) \end{equation*}$$ and $$\begin{equation*} HaK=(Ha)K. \end{equation*}$$ By definition, $$\begin{align*} H(aK) &=\{h\circ x:h\in H,\;x\in aK\}\\ &=\{h\circ(a\circ k):h\in H,\;k\in K\}\\ &=\{h\circ a\circ k:h\in H,\;k\in K\}\\ &=HaK. \end{align*}$$ Similarly, $$\begin{align*} (Ha)K &=\{y\circ k:y\in Ha,\;k\in K\}\\ &=\{(h\circ a)\circ k:h\in H,\;k\in K\}\\ &=\{h\circ a\circ k:h\in H,\;k\in K\}\\ &=HaK. \end{align*}$$ Hence, $$\begin{equation*} HaK=H(aK) \end{equation*}$$ and $$\begin{equation*} HaK=(Ha)K. \end{equation*}$$ $\square$ ::: :::example[Double Coset in an Additive Group] Let $G=\mathbb{Z}_{12}$ under addition modulo $12$. Let $$\begin{equation*} H=\{0,4,8\} \end{equation*}$$ and $$\begin{equation*} K=\{0,6\}. \end{equation*}$$ For $a=1$, the double coset becomes $$\begin{equation*} H+1+K=\{h+1+k:h\in H,\;k\in K\}. \end{equation*}$$ Computing all possible sums, we get $$\begin{align*} H+1+K &=\{0+1+0,\;0+1+6,\;4+1+0,\;4+1+6,\;8+1+0,\;8+1+6\}\\ &=\{1,7,5,11,9,15\}\\ &=\{1,5,7,9,11,3\}\\ &=\{1,3,5,7,9,11\}. \end{align*}$$ Thus the double coset contains six elements. ::: A double coset becomes concrete when we compute it inside the additive group of integers modulo $n$. In this model, the expression $H+a+K$ means that every element of $H$ is added to the representative $a$, and then every element of $K$ is added. Enter comma-separated elements for $H$ and $K$, then change the representative to see how the resulting subset changes. Notice that different pairs $(h,k)$ can give the same final element, so the raw list of sums may contain duplicates. This helps explain why double cosets are sets, not lists of representations. :::scientific-preview[Double Coset Explorer] ::: :::exercise[Exercises] 1. Let $G=\mathbb{Z}_{8}$, $H=\{0,4\}$, and $K=\{0,2,4,6\}$. Find $H+1+K$. 2. Let $(G,\circ)$ be a group and let $H,K$ be subgroups of $G$. Prove that $HaK\neq\varnothing$ for every $a\in G$. 3. Let $(G,\circ)$ be a group and let $H,K$ be subgroups of $G$. Prove that if $H=\{e\}$, then $HaK=aK$. ::: :::answer[Answers] 1. We have $$\begin{align*} H+1+K &=\{0+1+k:k\in K\}\cup\{4+1+k:k\in K\}\\ &=\{1,3,5,7\}\cup\{5,7,1,3\}\\ &=\{1,3,5,7\}. \end{align*}$$ 2. Since $e\in H$ and $e\in K$, we have $$\begin{equation*} a=e\circ a\circ e\in HaK. \end{equation*}$$ Therefore $HaK\neq\varnothing$. 3. If $H=\{e\}$, then $$\begin{align*} HaK &=\{e\circ a\circ k:k\in K\}\\ &=\{a\circ k:k\in K\}\\ &=aK. \end{align*}$$ ::: :::faq Q: What is a double coset? A: A double coset is a subset of the form $HaK=\{h\circ a\circ k:h\in H,\;k\in K\}$. Q: Is every double coset a subgroup? A: No. A double coset is a subset of $G$, but it need not be a subgroup. Q: How is a double coset related to an ordinary coset? A: If one of the two subgroups is the trivial subgroup, then the double coset reduces to an ordinary left or right coset. ::: :::call-to-action subtitle: Continue with the equivalence relation whose classes are precisely the double cosets. button: Next: Double Cosets as Equivalence Classes | published/abstract-algebra/introduction-to-groups/double-cosets/double-cosets-as-equivalence-classes.md :::