Having seen how abelian and cyclic properties pass to quotient groups, we now examine how subgroups behave inside a quotient. If $H\trianglelefteq G$ and $H\subseteq K\subseteq G$, then the cosets of $H$ lying inside $K$ form a subgroup $K/H$ of $G/H$. This is the first appearance of a correspondence principle: subgroups between $H$ and $G$ produce subgroups of $G/H$. The most important case is normality. A subgroup $K/H$ is normal in $G/H$ exactly when $K$ is normal in $G$. :::definition[Subgroup in a Quotient Group] Let $(G,\circ)$ be a group, let $H\trianglelefteq G$, and let $K$ be a subgroup of $G$ such that $H\subseteq K$. The subgroup determined by $K$ in $G/H$ is $$\begin{equation*} K/H=\{kH:k\in K\}. \end{equation*}$$ ::: :::theorem Let $(G,\circ)$ be a group, let $H\trianglelefteq G$, and let $K$ be a subgroup of $G$ such that $H\subseteq K\subseteq G$. Then $K/H$ is a subgroup of $G/H$. ::: :::proof Given that $(G,\circ)$ is a group, $H\trianglelefteq G$, and $K$ is a subgroup of $G$ such that $H\subseteq K\subseteq G$. To prove that $K/H$ is a subgroup of $G/H$. Since $H\trianglelefteq G$, the quotient group $G/H$ exists. Since $e\in K$, we get $$\begin{equation*} H=eH\in K/H. \end{equation*}$$ Therefore $K/H$ is non-empty. Let $aH,bH\in K/H$. Then $a,b\in K$. Since $K$ is a subgroup of $G$, we get $$\begin{equation*} a\circ b^{-1}\in K. \end{equation*}$$ Now the inverse of $bH$ in $G/H$ is $b^{-1}H$, and $$\begin{align*} (aH)(bH)^{-1} &=(aH)(b^{-1}H)\\ &=(a\circ b^{-1})H\\ &\in K/H. \end{align*}$$ By the one-step subgroup criterion, $K/H$ is a subgroup of $G/H$. Hence, $K/H\leq G/H$. $\square$ ::: :::theorem Let $(G,\circ)$ be a group, let $H,K$ be subgroups of $G$ such that $$\begin{equation*} H\subseteq K\subseteq G, \end{equation*}$$ and let $H\trianglelefteq G$. Then $K/H$ is normal in $G/H$ if and only if $K$ is normal in $G$. ::: :::proof Given that $(G,\circ)$ is a group, $H,K$ are subgroups of $G$, $H\subseteq K\subseteq G$, and $H\trianglelefteq G$. To prove that $K/H$ is normal in $G/H$ if and only if $K$ is normal in $G$. [1] Let $K/H$ be normal in $G/H$. To prove that $K$ is normal in $G$. Let $x\in G$ and let $k\in K$. Since $K/H$ is normal in $G/H$, $$\begin{equation*} (xH)(kH)(xH)^{-1}\in K/H. \end{equation*}$$ Thus there exists $k_1\in K$ such that $$\begin{equation*} (x\circ k\circ x^{-1})H=k_1H. \end{equation*}$$ Therefore $$\begin{equation*} k_1^{-1}\circ x\circ k\circ x^{-1}\in H. \end{equation*}$$ Since $H\subseteq K$ and $k_1\in K$, we get $$\begin{equation*} x\circ k\circ x^{-1}\in K. \end{equation*}$$ Therefore $$\begin{equation*} xKx^{-1}\subseteq K \quad \forall x\in G. \end{equation*}$$ Thus $K$ is normal in $G$. [2] Let $K$ be normal in $G$. To prove that $K/H$ is normal in $G/H$. Let $xH\in G/H$ and let $kH\in K/H$. Since $K$ is normal in $G$, $$\begin{equation*} x\circ k\circ x^{-1}\in K. \end{equation*}$$ Therefore $$\begin{align*} (xH)(kH)(xH)^{-1} &=(x\circ k\circ x^{-1})H\\ &\in K/H. \end{align*}$$ Thus $$\begin{equation*} (xH)(K/H)(xH)^{-1}\subseteq K/H \quad \forall xH\in G/H. \end{equation*}$$ Therefore $K/H$ is normal in $G/H$. Hence, $K/H\trianglelefteq G/H$ if and only if $K\trianglelefteq G$. $\square$ ::: :::example Let $G=(\mathbb{Z},+)$, let $H=6\mathbb{Z}$, and let $K=2\mathbb{Z}$. Then $$\begin{equation*} 6\mathbb{Z}\subseteq 2\mathbb{Z}\subseteq \mathbb{Z}. \end{equation*}$$ Since $\mathbb{Z}$ is abelian, both $H$ and $K$ are normal in $G$. Therefore $$\begin{equation*} K/H=2\mathbb{Z}/6\mathbb{Z} \end{equation*}$$ is a normal subgroup of $$\begin{equation*} \mathbb{Z}/6\mathbb{Z}. \end{equation*}$$ Hence, normality of $K$ in $G$ produces normality of $K/H$ in $G/H$. $\square$ ::: We observe the chain $H\subseteq K\subseteq G$ and its image $K/H\leq G/H$. Use the checkboxes to decide which hypotheses are available. The subgroup $K/H$ exists inside the quotient when $H$ is normal in $G$ and contained in $K$. Normality of $K/H$ inside $G/H$ is equivalent to normality of $K$ inside $G$. :::scientific-preview[Quotient Subgroup Correspondence Explorer] ::: The quotient $K/H$ keeps only the cosets coming from elements of $K$. The normality statement is not a new accident inside the quotient; it reflects the original normality of $K$ in $G$. :::solved-problem Let $H\trianglelefteq G$ and let $K$ be a subgroup of $G$ with $H\subseteq K$. Prove that $K/H$ contains the identity element of $G/H$. ::: :::solution Since $K$ is a subgroup of $G$, we have $e\in K$. Therefore $$\begin{equation*} eH=H\in K/H. \end{equation*}$$ The identity element of $G/H$ is $H$. Thus $K/H$ contains the identity element of $G/H$. ::: :::solved-problem Let $H\trianglelefteq G$, $H\subseteq K\subseteq G$, and $K\trianglelefteq G$. Prove that $K/H\trianglelefteq G/H$. ::: :::solution Let $xH\in G/H$ and let $kH\in K/H$. Since $K\trianglelefteq G$, $$\begin{equation*} x\circ k\circ x^{-1}\in K. \end{equation*}$$ Therefore $$\begin{align*} (xH)(kH)(xH)^{-1} &=(x\circ k\circ x^{-1})H\\ &\in K/H. \end{align*}$$ Thus $$\begin{equation*} K/H\trianglelefteq G/H. \end{equation*}$$ ::: When $G$, $K$, and $H$ are finite with $H\subseteq K$, the subgroup quotient $K/H$ has order $|K|/|H|$. Enter compatible orders to compute both $|G/H|$ and $|K/H|$. This numerical check reinforces the containment picture: $K/H$ is a smaller subgroup sitting inside $G/H$. :::calculator[Subgroup Quotient Order Calculator] ::: :::exercise 1. Define $K/H$ when $H\trianglelefteq G$ and $H\subseteq K\subseteq G$. 2. Prove that $K/H$ is a subgroup of $G/H$. 3. Prove that if $K\trianglelefteq G$, then $K/H\trianglelefteq G/H$. 4. Prove the converse: if $K/H\trianglelefteq G/H$, then $K\trianglelefteq G$. 5. Give an example of $H\subseteq K\subseteq G$ using subgroups of $\mathbb{Z}$. ::: :::answer 1. $K/H=\{kH:k\in K\}$. 2. If $aH,bH\in K/H$, then $(aH)(bH)^{-1}=(ab^{-1})H\in K/H$ because $ab^{-1}\in K$. 3. If $K\trianglelefteq G$, then conjugating $kH$ by $xH$ gives $(xkx^{-1})H\in K/H$. 4. If $K/H\trianglelefteq G/H$, then $(xH)(kH)(xH)^{-1}\in K/H$, forcing $xkx^{-1}\in K$. 5. One example is $6\mathbb{Z}\subseteq 2\mathbb{Z}\subseteq \mathbb{Z}$. ::: :::faq[Frequently Asked Questions] Q: Why must $H$ be contained in $K$? A: The expression $K/H$ uses cosets of $H$ inside $K$, so $H\subseteq K$ is required. Q: Is $K/H$ always normal in $G/H$? A: No. It is normal in $G/H$ exactly when $K$ is normal in $G$. Q: What is the identity element of $K/H$? A: The identity element is $H=eH$, the same identity coset used in $G/H$. ::: :::call-to-action[Continue to the Order of a Quotient Group] subtitle: Use index and Lagrange's theorem to count elements in finite quotient groups. button: Next Lesson | published/abstract-algebra/subgroups-and-normal-subgroups/quotient-groups/order-of-a-quotient-group.md button-ghost: Previous Lesson | published/abstract-algebra/subgroups-and-normal-subgroups/quotient-groups/abelian-and-cyclic-quotient-groups.md :::