We now prove that all cosets of the same subgroup have the same number of elements. This fact is not only a counting statement; it is the mechanism behind Lagrange's theorem. Each coset is a translated copy of the subgroup, so it has the same size as the subgroup. We also prove that the collection of left cosets and the collection of right cosets have the same cardinality. This does not mean each left coset equals a right coset, but it does mean they can be paired perfectly. :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Then any two left cosets of $H$ in $G$ have the same cardinality. ::: :::proof Given that $(G,\circ)$ is a group and $H$ is a subgroup of $G$. To prove that any two left cosets of $H$ in $G$ have the same cardinality. Let $a,b\in G$. We prove that $aH$ and $bH$ have the same cardinality. Define $f:aH\to bH$ by $$\begin{equation*} f(a\circ h)=b\circ h \end{equation*}$$ for every $h\in H$. First, we prove that $f$ is well-defined. Let $a\circ h_1=a\circ h_2$, where $h_1,h_2\in H$. Then $$\begin{align*} a\circ h_1=a\circ h_2 &\implies a^{-1}\circ a\circ h_1=a^{-1}\circ a\circ h_2\\ &\implies h_1=h_2. \end{align*}$$ Therefore $b\circ h_1=b\circ h_2$, so $f$ is well-defined. To prove that $f$ is one-one, let $f(a\circ h_1)=f(a\circ h_2)$. Then $$\begin{align*} f(a\circ h_1)=f(a\circ h_2) &\implies b\circ h_1=b\circ h_2\\ &\implies b^{-1}\circ b\circ h_1=b^{-1}\circ b\circ h_2\\ &\implies h_1=h_2\\ &\implies a\circ h_1=a\circ h_2. \end{align*}$$ Therefore $f$ is one-one. To prove that $f$ is onto, let $y\in bH$. Then there exists $h\in H$ such that $$\begin{equation*} y=b\circ h. \end{equation*}$$ Now $a\circ h\in aH$ and $$\begin{equation*} f(a\circ h)=b\circ h=y. \end{equation*}$$ Therefore $f$ is onto. Hence, $aH$ and $bH$ have the same cardinality. $\square$ ::: :::corollary Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $a\in G$, then $$\begin{equation*} |aH|=|H|. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $a\in G$. To prove that $|aH|=|H|$. Since $H=eH$, the previous theorem gives $$\begin{equation*} |aH|=|eH|. \end{equation*}$$ Since $eH=H$, we get $$\begin{equation*} |aH|=|H|. \end{equation*}$$ Hence, $|aH|=|H|$. $\square$ ::: We observe that translating a subgroup does not change the number of elements. Choose a subgroup of $\mathbb{Z}_n$ and a representative, and the preview pairs each element of $H$ with the corresponding element of $a+H$. The values may change, but the number of entries stays the same. This is the finite counting idea behind Lagrange's theorem. :::scientific-preview[Coset Cardinality Explorer] ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Then any two right cosets of $H$ in $G$ have the same cardinality. ::: :::proof Given that $(G,\circ)$ is a group and $H$ is a subgroup of $G$. To prove that any two right cosets of $H$ in $G$ have the same cardinality. Let $a,b\in G$. We prove that $Ha$ and $Hb$ have the same cardinality. Define $g:Ha\to Hb$ by $$\begin{equation*} g(h\circ a)=h\circ b \end{equation*}$$ for every $h\in H$. First, we prove that $g$ is well-defined. Let $h_1\circ a=h_2\circ a$, where $h_1,h_2\in H$. Then $$\begin{align*} h_1\circ a=h_2\circ a &\implies h_1\circ a\circ a^{-1}=h_2\circ a\circ a^{-1}\\ &\implies h_1=h_2. \end{align*}$$ Therefore $h_1\circ b=h_2\circ b$, so $g$ is well-defined. To prove that $g$ is one-one, let $g(h_1\circ a)=g(h_2\circ a)$. Then $$\begin{align*} g(h_1\circ a)=g(h_2\circ a) &\implies h_1\circ b=h_2\circ b\\ &\implies h_1\circ b\circ b^{-1}=h_2\circ b\circ b^{-1}\\ &\implies h_1=h_2\\ &\implies h_1\circ a=h_2\circ a. \end{align*}$$ Therefore $g$ is one-one. To prove that $g$ is onto, let $y\in Hb$. Then there exists $h\in H$ such that $$\begin{equation*} y=h\circ b. \end{equation*}$$ Now $h\circ a\in Ha$ and $$\begin{equation*} g(h\circ a)=h\circ b=y. \end{equation*}$$ Therefore $g$ is onto. Hence, $Ha$ and $Hb$ have the same cardinality. $\square$ ::: :::corollary Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $a\in G$, then $$\begin{equation*} |Ha|=|H|. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $a\in G$. To prove that $|Ha|=|H|$. Since $H=He$, the previous theorem gives $$\begin{equation*} |Ha|=|He|. \end{equation*}$$ Since $He=H$, we get $$\begin{equation*} |Ha|=|H|. \end{equation*}$$ Hence, $|Ha|=|H|$. $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Then there is a one-one correspondence from the set of all left cosets of $H$ in $G$ onto the set of all right cosets of $H$ in $G$. ::: :::proof Given that $(G,\circ)$ is a group and $H$ is a subgroup of $G$. To prove that there is a one-one correspondence from the set of all left cosets of $H$ onto the set of all right cosets of $H$. Let $\mathcal{L}$ be the set of all left cosets of $H$ and let $\mathcal{R}$ be the set of all right cosets of $H$. Define $\Phi:\mathcal{L}\to \mathcal{R}$ by $$\begin{equation*} \Phi(aH)=Ha^{-1}. \end{equation*}$$ First, we prove that $\Phi$ is well-defined. Let $aH=bH$. By the left coset equality criterion, $$\begin{equation*} b^{-1}\circ a\in H. \end{equation*}$$ For right cosets, we have $$\begin{align*} Ha^{-1}=Hb^{-1} &\iff b^{-1}\circ a\in H. \end{align*}$$ Therefore $Ha^{-1}=Hb^{-1}$. Hence, $\Phi$ is well-defined. To prove that $\Phi$ is one-one, let $$\begin{equation*} \Phi(aH)=\Phi(bH). \end{equation*}$$ Then $$\begin{equation*} Ha^{-1}=Hb^{-1}. \end{equation*}$$ By the right coset equality criterion, $$\begin{equation*} b^{-1}\circ a\in H. \end{equation*}$$ By the left coset equality criterion, $$\begin{equation*} aH=bH. \end{equation*}$$ Therefore $\Phi$ is one-one. To prove that $\Phi$ is onto, let $Ha\in \mathcal{R}$. Then $a^{-1}H\in \mathcal{L}$ and $$\begin{align*} \Phi(a^{-1}H) &=H(a^{-1})^{-1}\\ &=Ha. \end{align*}$$ Therefore $\Phi$ is onto. Hence, there is a one-one correspondence from the set of all left cosets of $H$ onto the set of all right cosets of $H$. $\square$ ::: :::example[Cardinality of Cosets in Integers Modulo Twelve] Let $G=\mathbb{Z}_{12}$ and let $H=\{0,4,8\}$. Each coset of $H$ has $3$ elements: $$\begin{align*} 0+H&=\{0,4,8\},\\ 1+H&=\{1,5,9\},\\ 2+H&=\{2,6,10\},\\ 3+H&=\{3,7,11\}. \end{align*}$$ Therefore every coset has the same cardinality as $H$. ::: :::solved-problem[Counting Elements in a Coset] Let $(G,\circ)$ be a group and let $H$ be a finite subgroup of $G$ with $|H|=6$. If $a\in G$, find $|aH|$ and $|Ha|$. ::: :::solution[Solution] Let $(G,\circ)$ be a group and let $H$ be a finite subgroup of $G$ with $|H|=6$. Since every left coset has the same cardinality as $H$, we get $$\begin{equation*} |aH|=|H|=6. \end{equation*}$$ Since every right coset has the same cardinality as $H$, we get $$\begin{equation*} |Ha|=|H|=6. \end{equation*}$$ ::: We compute the size of a modular subgroup and one of its cosets. The list of elements changes after translation, but the number of elements does not. This is because the map $h\mapsto a+h$ pairs every subgroup element with exactly one coset element. The calculator verifies the equality $|a+H|=|H|$ in concrete examples. :::calculator[Coset Size Calculator] ::: :::exercise[Exercises] 1. Let $H$ be a finite subgroup of a group $G$ and let $|H|=4$. Find the number of elements in $aH$ for any $a\in G$. 2. Let $G=\mathbb{Z}_{18}$ and let $H=\{0,6,12\}$. Find the number of elements in each coset of $H$. 3. Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Describe a one-one correspondence from left cosets of $H$ to right cosets of $H$. ::: :::answer[Answers] 1. We have $$\begin{equation*} |aH|=|H|=4. \end{equation*}$$ 2. Since $|H|=3$, each coset of $H$ has $3$ elements. 3. The correspondence is $$\begin{equation*} aH\mapsto Ha^{-1}. \end{equation*}$$ ::: :::faq Q: Does every coset have the same number of elements as the subgroup? A: Yes. Both left and right cosets have the same cardinality as $H$. Q: Does equal cardinality mean the coset is equal to $H$? A: No. A coset may have the same size as $H$ without being equal to $H$. Q: Are the numbers of left and right cosets the same? A: Yes. The map $aH\mapsto Ha^{-1}$ gives a one-one correspondence. ::: :::call-to-action subtitle: Continue with coset formulas involving intersections of subgroups. button: Next: Cosets and Intersections of Subgroups | published/abstract-algebra/introduction-to-groups/cosets-of-subgroups/cosets-and-intersections-of-subgroups.md :::