After proving that cosets partition a group, we now need efficient tests for deciding whether two elements determine the same coset. Instead of listing all elements, we can use a product involving inverses. For left cosets, the key expression is $a^{-1}\circ b$. For right cosets, the key expression is $b\circ a^{-1}$. These criteria are practical tools: they turn questions about subsets into questions about whether a single element lies in the subgroup. :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $a,b\in G$, then $$\begin{equation*} b\in aH \iff a^{-1}\circ b\in H. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $a,b\in G$. To prove that $b\in aH \iff a^{-1}\circ b\in H$. [1] Let $b\in aH$. Then there exists $h\in H$ such that $$\begin{equation*} b=a\circ h. \end{equation*}$$ Therefore $$\begin{align*} b=a\circ h &\implies a^{-1}\circ b=a^{-1}\circ a\circ h\\ &\implies a^{-1}\circ b=e\circ h\\ &\implies a^{-1}\circ b=h. \end{align*}$$ Since $h\in H$, we get $a^{-1}\circ b\in H$. [2] Let $a^{-1}\circ b\in H$. Then $$\begin{align*} b &=a\circ (a^{-1}\circ b). \end{align*}$$ Since $a^{-1}\circ b\in H$, we get $b\in aH$. Hence, $$\begin{equation*} b\in aH \iff a^{-1}\circ b\in H. \end{equation*}$$ $\square$ ::: We observe the left coset membership test in additive notation. Choose $a$, $b$, and a subgroup of $\mathbb{Z}_n$, then compare membership of $b$ in $a+H$ with membership of $-a+b$ in $H$. The two answers agree because subtracting $a$ from $b$ isolates the subgroup element that would produce $b$. This turns a set-membership question into one modular calculation. :::scientific-preview[Membership Criterion Explorer] ::: :::corollary Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $a,b\in G$, then $$\begin{equation*} aH=bH \iff a^{-1}\circ b\in H. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $a,b\in G$. To prove that $aH=bH \iff a^{-1}\circ b\in H$. [1] Let $aH=bH$. Since $e\in H$, we get $$\begin{align*} b &=b\circ e\\ &\in bH\\ &=aH. \end{align*}$$ By the membership criterion for left cosets, $$\begin{equation*} b\in aH \implies a^{-1}\circ b\in H. \end{equation*}$$ Therefore $a^{-1}\circ b\in H$. [2] Let $a^{-1}\circ b\in H$. By the membership criterion for left cosets, we get $b\in aH$. Since $b\in bH$, we get $$\begin{equation*} aH\cap bH\neq\varnothing. \end{equation*}$$ By the equal-or-disjoint theorem for left cosets, $$\begin{equation*} aH=bH. \end{equation*}$$ Hence, $$\begin{equation*} aH=bH \iff a^{-1}\circ b\in H. \end{equation*}$$ $\square$ ::: :::corollary Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $a,b\in G$, then $$\begin{equation*} aH=bH \iff b^{-1}\circ a\in H. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $a,b\in G$. To prove that $aH=bH \iff b^{-1}\circ a\in H$. By the previous corollary, $$\begin{equation*} aH=bH \iff a^{-1}\circ b\in H. \end{equation*}$$ Since $H$ is a subgroup of $G$, an element belongs to $H$ if and only if its inverse belongs to $H$. Therefore $$\begin{align*} a^{-1}\circ b\in H &\iff (a^{-1}\circ b)^{-1}\in H\\ &\iff b^{-1}\circ a\in H. \end{align*}$$ Hence, $$\begin{equation*} aH=bH \iff b^{-1}\circ a\in H. \end{equation*}$$ $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $a,b\in G$, then $$\begin{equation*} b\in Ha \iff b\circ a^{-1}\in H. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $a,b\in G$. To prove that $b\in Ha \iff b\circ a^{-1}\in H$. [1] Let $b\in Ha$. Then there exists $h\in H$ such that $$\begin{equation*} b=h\circ a. \end{equation*}$$ Therefore $$\begin{align*} b=h\circ a &\implies b\circ a^{-1}=h\circ a\circ a^{-1}\\ &\implies b\circ a^{-1}=h\circ e\\ &\implies b\circ a^{-1}=h. \end{align*}$$ Since $h\in H$, we get $b\circ a^{-1}\in H$. [2] Let $b\circ a^{-1}\in H$. Then $$\begin{align*} b &=(b\circ a^{-1})\circ a. \end{align*}$$ Since $b\circ a^{-1}\in H$, we get $b\in Ha$. Hence, $$\begin{equation*} b\in Ha \iff b\circ a^{-1}\in H. \end{equation*}$$ $\square$ ::: :::corollary Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $a,b\in G$, then $$\begin{equation*} Ha=Hb \iff b\circ a^{-1}\in H. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $a,b\in G$. To prove that $Ha=Hb \iff b\circ a^{-1}\in H$. [1] Let $Ha=Hb$. Since $e\in H$, we get $$\begin{align*} b &=e\circ b\\ &\in Hb\\ &=Ha. \end{align*}$$ By the membership criterion for right cosets, $$\begin{equation*} b\in Ha \implies b\circ a^{-1}\in H. \end{equation*}$$ Therefore $b\circ a^{-1}\in H$. [2] Let $b\circ a^{-1}\in H$. By the membership criterion for right cosets, we get $b\in Ha$. Since $b\in Hb$, we get $$\begin{equation*} Ha\cap Hb\neq\varnothing. \end{equation*}$$ By the equal-or-disjoint theorem for right cosets, $$\begin{equation*} Ha=Hb. \end{equation*}$$ Hence, $$\begin{equation*} Ha=Hb \iff b\circ a^{-1}\in H. \end{equation*}$$ $\square$ ::: :::example[Equality Test in Additive Notation] Let $G=\mathbb{Z}_{12}$ under addition modulo $12$ and let $H=\{0,4,8\}$. We test whether $1+H=5+H$. In additive notation, the left coset equality criterion becomes $$\begin{equation*} 1+H=5+H \iff -1+5\in H. \end{equation*}$$ Now $$\begin{align*} -1+5 &=4\\ &\in H. \end{align*}$$ Therefore $$\begin{equation*} 1+H=5+H. \end{equation*}$$ ::: :::solved-problem[A Membership Test] Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Suppose $a^{-1}\circ b\notin H$. Prove that $b\notin aH$. ::: :::solution[Solution] Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Given that $a^{-1}\circ b\notin H$. By the left coset membership criterion, $$\begin{equation*} b\in aH \iff a^{-1}\circ b\in H. \end{equation*}$$ Since $a^{-1}\circ b\notin H$, we get $$\begin{equation*} b\notin aH. \end{equation*}$$ ::: We test whether two modular cosets are equal using the additive criterion. Enter $a$ and $b$, then compute $-a+b$ modulo $n$ and check whether that value lies in $H$. If it does, the two representatives determine the same coset. This avoids listing every element first, although the calculator also displays the two cosets for verification. :::calculator[Coset Equality Criterion Calculator] ::: :::exercise[Exercises] 1. Let $G=\mathbb{Z}_{15}$ and let $H=\{0,5,10\}$. Decide whether $2+H=12+H$. 2. Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Prove that $aH=bH$ if and only if $b\in aH$. 3. Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Prove that $Ha=Hb$ if and only if $b\in Ha$. ::: :::answer[Answers] 1. In additive notation, $$\begin{align*} -2+12 &=10\\ &\in H. \end{align*}$$ Therefore $2+H=12+H$. 2. If $aH=bH$, then $b\in bH=aH$. Conversely, if $b\in aH$, then $aH\cap bH\neq\varnothing$, so $aH=bH$. 3. If $Ha=Hb$, then $b\in Hb=Ha$. Conversely, if $b\in Ha$, then $Ha\cap Hb\neq\varnothing$, so $Ha=Hb$. ::: :::faq Q: Why does the left coset criterion use $a^{-1}\circ b$? A: Because $b\in aH$ means $b=a\circ h$ for some $h\in H$, so multiplying by $a^{-1}$ on the left isolates $h$. Q: Why does the right coset criterion use $b\circ a^{-1}$? A: Because $b\in Ha$ means $b=h\circ a$, so multiplying by $a^{-1}$ on the right isolates $h$. Q: Can we interchange these tests in a nonabelian group? A: No. The order of multiplication matters in a nonabelian group. ::: :::call-to-action subtitle: Continue with the equivalence relations whose classes are cosets. button: Next: Cosets as Equivalence Classes | published/abstract-algebra/introduction-to-groups/cosets-of-subgroups/cosets-as-equivalence-classes.md :::