The isomorphism theorems show how quotients and images are connected. We now organize all subgroups involved in an epimorphism. The focus keyword correspondence theorem states that if $f:G\to G_1$ is onto, then subgroups of $G_1$ correspond exactly to subgroups of $G$ that contain $\ker f$. In the quotient case, subgroups of $G/N$ correspond exactly to subgroups of $G$ containing $N$. This theorem is a map of the subgroup structure after quotienting. Students often try to list subgroups of a quotient directly, but correspondence gives a systematic method. :::theorem[Correspondence Theorem] Let $f:G\to G_1$ be an epimorphism and let $N=\ker f$. Then there is a one-to-one inclusion-preserving correspondence between $$\begin{equation*} \{H\leq G\mid N\subseteq H\} \end{equation*}$$ and $$\begin{equation*} \{K\leq G_1\}. \end{equation*}$$ The correspondence is $H\mapsto f(H)$, and its inverse is $K\mapsto f^{-1}(K)$. Moreover, $H\trianglelefteq G$ if and only if $f(H)\trianglelefteq G_1$. ::: :::proof Given that $f:G\to G_1$ is an epimorphism and $N=\ker f$. To prove the one-to-one inclusion-preserving correspondence and the normality statement. Let $H\leq G$ with $N\subseteq H$. Since the image of a subgroup under a homomorphism is a subgroup, $f(H)\leq G_1$. Let $K\leq G_1$. Since inverse images of subgroups under homomorphisms are subgroups, $f^{-1}(K)\leq G$. Also, $e_1\in K$, so every element of $N$ maps into $K$. Hence $N\subseteq f^{-1}(K)$. Now prove that the two assignments are inverse. Let $H\leq G$ and $N\subseteq H$. First, $H\subseteq f^{-1}(f(H))$. Conversely, let $x\in f^{-1}(f(H))$. Then $f(x)\in f(H)$, so there exists $h\in H$ such that $f(x)=f(h)$. Hence $$\begin{align*} f(h^{-1}x)&=f(h)^{-1}f(x)\\ &=e_1. \end{align*}$$ Thus $h^{-1}x\in N\subseteq H$. Since $h\in H$, we get $x=h(h^{-1}x)\in H$. Therefore $f^{-1}(f(H))=H$. Let $K\leq G_1$. Since $f$ is onto, every $k\in K$ has the form $k=f(x)$ for some $x\in G$. Since $k\in K$, $x\in f^{-1}(K)$. Therefore $k\in f(f^{-1}(K))$. Thus $f(f^{-1}(K))=K$. The correspondence is inclusion-preserving because $H_1\subseteq H_2$ implies $f(H_1)\subseteq f(H_2)$, and $K_1\subseteq K_2$ implies $f^{-1}(K_1)\subseteq f^{-1}(K_2)$. If $H\trianglelefteq G$, then $f(H)\trianglelefteq G_1$ because $f$ is onto. Conversely, if $K\trianglelefteq G_1$, then $f^{-1}(K)\trianglelefteq G$. Hence normality is preserved under the correspondence. $\square$ ::: :::corollary[Subgroups of a Quotient Group] Let $N\trianglelefteq G$. Then every subgroup of $G/N$ has the form $H/N$, where $N\subseteq H\leq G$. Moreover, $$\begin{equation*} H/N\trianglelefteq G/N \end{equation*}$$ if and only if $$\begin{equation*} H\trianglelefteq G. \end{equation*}$$ ::: :::proof Given that $N\trianglelefteq G$. To prove the subgroup and normal subgroup correspondence for $G/N$. Let $\pi:G\to G/N$ be the natural homomorphism, $\pi(a)=aN$. Then $\ker\pi=N$. Applying the correspondence theorem to $\pi$, the subgroups of $G/N$ correspond exactly to subgroups $H$ of $G$ such that $N\subseteq H$. The image of such an $H$ is $$\begin{equation*} \pi(H)=H/N. \end{equation*}$$ Thus every subgroup of $G/N$ has the form $H/N$ with $N\subseteq H\leq G$. The normality statement follows from the normality-preserving part of the correspondence theorem. $\square$ ::: For quotient groups, the correspondence theorem is often the fastest way to list subgroups. Instead of guessing subgroups of $G/N$, list the subgroups of $G$ that contain $N$, and then divide each of them by $N$. This works because the natural homomorphism $G\to G/N$ has kernel $N$. The theorem also says that normal subgroups of the quotient come from normal subgroups of the original group that contain $N$. :::example[Correspondence for Z to Z Twelve] Let $f:(\mathbb{Z},+)\to(\mathbb{Z}_{12},+_{12})$ be defined by $f(n)=[n]_{12}$. Then $\ker f=(12)$. The subgroups of $\mathbb{Z}$ containing $(12)$ are $$\begin{equation*} (12),\quad (6),\quad (4),\quad (3),\quad (2),\quad \mathbb{Z}. \end{equation*}$$ Their images are $$\begin{equation*} \langle[0]\rangle,\quad \langle[6]\rangle,\quad \langle[4]\rangle,\quad \langle[3]\rangle,\quad \langle[2]\rangle,\quad \mathbb{Z}_{12}. \end{equation*}$$ These are exactly the subgroups of $\mathbb{Z}_{12}$. ::: This preview shows the quotient version of the correspondence theorem for $\mathbb Z\to\mathbb Z_n$. Choose a modulus $n$ and compare each subgroup $(d)$ containing $(n)$ with its image $\langle[d]\rangle$ in $\mathbb Z_n$. The rows are ordered from the largest subgroup inside $\mathbb Z$ down to the smallest containing $(n)$. Notice that inclusion is preserved by the correspondence, even though the notation for integer subgroups can feel reversed. Try $n=12$ first, then switch to $18$ to match the exercise. :::scientific-preview[Subgroup Correspondence Explorer] ::: The number of rows equals the number of divisors of $n$, which is also the number of subgroups of the cyclic group $\mathbb Z_n$. The table shows why the quotient method is systematic: every subgroup in the quotient is the image of exactly one subgroup above the kernel. :::solved-problem Let $N\trianglelefteq G$. Prove that every subgroup of $G/N$ is of the form $H/N$ for some $H$ satisfying $N\subseteq H\leq G$. ::: :::solution Let $N\trianglelefteq G$, and let $L\leq G/N$. Let $\pi:G\to G/N$ be the natural homomorphism. Define $$\begin{equation*} H=\pi^{-1}(L). \end{equation*}$$ Since inverse images of subgroups are subgroups, $H\leq G$. Since $\ker\pi=N$ and $e_{G/N}\in L$, we have $N\subseteq H$. Also, $$\begin{equation*} \pi(H)=L \end{equation*}$$ because $L$ is contained in the codomain of the onto map $\pi$. But $\pi(H)=H/N$. Therefore $$\begin{equation*} L=H/N. \end{equation*}$$ Hence every subgroup of $G/N$ has the required form. $\square$ ::: :::solved-problem Let $N\trianglelefteq G$ and $N\subseteq H\leq G$. Prove that $H/N\trianglelefteq G/N$ if and only if $H\trianglelefteq G$. ::: :::solution Let $N\trianglelefteq G$ and $N\subseteq H\leq G$. Let $\pi:G\to G/N$ be the natural homomorphism. Since $\ker\pi=N$, the correspondence theorem says that normal subgroups of $G/N$ correspond exactly to normal subgroups of $G$ containing $N$. Under this correspondence, $H$ corresponds to $$\begin{equation*} \pi(H)=H/N. \end{equation*}$$ Therefore $H/N\trianglelefteq G/N$ if and only if $H\trianglelefteq G$. $\square$ ::: :::tip When using the correspondence theorem for $G/N$, never list all subgroups of $G$; list only those that contain $N$. Subgroups not containing $N$ do not appear as subgroups of the quotient. ::: :::exercise 1. Use the correspondence theorem to list subgroups of $\mathbb{Z}_{18}$. 2. If $G/N$ is simple, what can be said about normal subgroups of $G$ containing $N$? 3. If $H_1\subseteq H_2$ and both contain $\ker f$, what happens to their images under $f$? ::: :::answer 1. Subgroups correspond to divisors of $18$: $\langle[0]\rangle$, $\langle[9]\rangle$, $\langle[6]\rangle$, $\langle[3]\rangle$, $\langle[2]\rangle$, and $\mathbb{Z}_{18}$. 2. The only normal subgroups of $G$ containing $N$ are $N$ and $G$. 3. Inclusion is preserved, so $f(H_1)\subseteq f(H_2)$. ::: :::faq[Frequently Asked Questions] Q: Why must $f$ be onto in the correspondence theorem? A: Onto-ness ensures every subgroup element in $G_1$ has a preimage and makes $f(f^{-1}(K))=K$. Q: What is the quotient version of the theorem? A: Subgroups of $G/N$ correspond to subgroups of $G$ containing $N$. Q: Does the theorem preserve normality? A: Yes. Normal subgroups correspond to normal subgroups under the stated correspondence. ::: :::call-to-action[Review the Isomorphism Theorems] subtitle: Revisit factorization, the first theorem, homomorphic images, and quotient-by-stages as one connected framework. button: Return to Factorization | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/isomorphism-and-correspondence-theorems/factorization-through-quotients button-ghost: Review the Third Isomorphism Theorem | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/isomorphism-and-correspondence-theorems/third-isomorphism-theorem :::