The previous lecture introduced left and right cosets. We now study the first important special case: what happens when the representative itself belongs to the subgroup. This case is foundational because it tells us exactly when a coset is not new at all. If the representative lies inside $H$, then multiplying $H$ by that representative simply reproduces $H$. Conversely, if a coset determined by $h$ equals $H$, then $h$ must lie in $H$. This result is often the first place where students see that cosets behave like shifted copies of a subgroup. :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $h\in G$, then $$\begin{equation*} h\in H \iff hH=H. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $h\in G$. To prove that $h\in H \iff hH=H$. [1] Let $h\in H$. To prove that $hH=H$. Let $x\in hH$. Then there exists $u\in H$ such that $$\begin{equation*} x=h\circ u. \end{equation*}$$ Since $h,u\in H$ and $H$ is closed under $\circ$, we get $x\in H$. Therefore $$\begin{equation*} hH\subseteq H. \end{equation*}$$ Let $y\in H$. Since $h\in H$ and $H$ is a subgroup of $G$, we get $h^{-1}\in H$. Therefore $h^{-1}\circ y\in H$. Now $$\begin{align*} y &=h\circ (h^{-1}\circ y). \end{align*}$$ Therefore $y\in hH$. Therefore $$\begin{equation*} H\subseteq hH. \end{equation*}$$ Therefore $$\begin{equation*} hH=H. \end{equation*}$$ [2] Let $hH=H$. To prove that $h\in H$. Since $e\in H$, we get $$\begin{align*} h &=h\circ e\\ &\in hH. \end{align*}$$ Since $hH=H$, we get $h\in H$. Hence, $$\begin{equation*} h\in H \iff hH=H. \end{equation*}$$ $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $h\in G$, then $$\begin{equation*} h\in H \iff Hh=H. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $h\in G$. To prove that $h\in H \iff Hh=H$. [1] Let $h\in H$. To prove that $Hh=H$. Let $x\in Hh$. Then there exists $u\in H$ such that $$\begin{equation*} x=u\circ h. \end{equation*}$$ Since $u,h\in H$ and $H$ is closed under $\circ$, we get $x\in H$. Therefore $$\begin{equation*} Hh\subseteq H. \end{equation*}$$ Let $y\in H$. Since $h\in H$ and $H$ is a subgroup of $G$, we get $h^{-1}\in H$. Therefore $y\circ h^{-1}\in H$. Now $$\begin{align*} y &=(y\circ h^{-1})\circ h. \end{align*}$$ Therefore $y\in Hh$. Therefore $$\begin{equation*} H\subseteq Hh. \end{equation*}$$ Therefore $$\begin{equation*} Hh=H. \end{equation*}$$ [2] Let $Hh=H$. To prove that $h\in H$. Since $e\in H$, we get $$\begin{align*} h &=e\circ h\\ &\in Hh. \end{align*}$$ Since $Hh=H$, we get $h\in H$. Hence, $$\begin{equation*} h\in H \iff Hh=H. \end{equation*}$$ $\square$ ::: We observe the special case where the representative already belongs to the subgroup. Choose a subgroup of $\mathbb{Z}_n$ and a representative, then compare $a+H$ with $H$. When $a$ lies in $H$, the translation only rearranges the same elements, so the coset is not new. When $a$ does not lie in $H$, the coset is a shifted copy that differs from $H$. :::scientific-preview[Coset Equals Subgroup Explorer] ::: :::corollary Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $a\in G-H$, then $$\begin{equation*} aH\neq H \end{equation*}$$ and $$\begin{equation*} Ha\neq H. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $a\in G-H$. To prove that $aH\neq H$ and $Ha\neq H$. Since $a\in G-H$, we get $a\notin H$. If possible, let $aH=H$. By the previous theorem on left cosets, $$\begin{equation*} aH=H \implies a\in H. \end{equation*}$$ A contradiction. Therefore $aH\neq H$. If possible, let $Ha=H$. By the previous theorem on right cosets, $$\begin{equation*} Ha=H \implies a\in H. \end{equation*}$$ A contradiction. Hence, $aH\neq H$ and $Ha\neq H$. $\square$ ::: :::example[Cosets Determined by Elements of a Subgroup] Let $G=\mathbb{Z}_{12}$ under addition modulo $12$ and let $$\begin{equation*} H=\{0,3,6,9\}. \end{equation*}$$ Since $3\in H$, the coset determined by $3$ is $$\begin{align*} 3+H &=\{3+0,\;3+3,\;3+6,\;3+9\}\\ &=\{3,\;6,\;9,\;12\}\\ &=\{0,3,6,9\}\\ &=H. \end{align*}$$ This example illustrates the theorem in additive notation. ::: :::solved-problem[Testing Whether a Coset Is the Subgroup] Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Suppose $aH=H$. Prove that $a^{-1}\in H$. ::: :::solution[Solution] Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Given that $aH=H$. Since $aH=H$, by the left coset criterion we get $$\begin{equation*} a\in H. \end{equation*}$$ Since $H$ is a subgroup of $G$, it is closed under taking inverses. Therefore $$\begin{equation*} a^{-1}\in H. \end{equation*}$$ ::: We test the theorem numerically in $\mathbb{Z}_n$. Enter a subgroup step and a representative, and the calculator checks both statements: whether the representative belongs to the subgroup and whether the coset equals the subgroup. Notice that the two answers always agree in this additive setting. This is the finite modular version of $h\in H$ if and only if $hH=H$. :::calculator[Coset Subgroup Equality Tester] ::: :::exercise[Exercises] 1. Let $G=\mathbb{Z}_{15}$ under addition modulo $15$ and let $H=\{0,5,10\}$. Find $10+H$. 2. Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $Ha=H$, prove that $a^{-1}\in H$. 3. Let $H$ be a subgroup of $G$. Prove that if $a\notin H$, then $aH$ cannot be equal to $H$. ::: :::answer[Answers] 1. Since $10\in H$, we get $$\begin{equation*} 10+H=H=\{0,5,10\}. \end{equation*}$$ 2. Since $Ha=H$, the right coset criterion gives $a\in H$. Since $H$ is a subgroup, $a^{-1}\in H$. 3. If $aH=H$, then the left coset criterion gives $a\in H$, which contradicts $a\notin H$. Therefore $aH\neq H$. ::: :::faq Q: Why does $hH=H$ when $h\in H$? A: Multiplication by an element already inside $H$ only rearranges the elements of $H$. Q: Is the condition $h\in H$ necessary? A: Yes. The theorem proves both directions: $hH=H$ happens exactly when $h\in H$. Q: Does the same result hold for right cosets? A: Yes. We also have $h\in H \iff Hh=H$. ::: :::call-to-action subtitle: Continue with the theorem that two cosets are either equal or disjoint. button: Next: Equal and Disjoint Cosets | published/abstract-algebra/introduction-to-groups/cosets-of-subgroups/equal-and-disjoint-cosets.md :::