The first isomorphism theorem shows that every epimorphism from $G$ produces a quotient of $G$. The focus keyword homomorphic images and normal subgroups captures this exact connection: homomorphic images of a group are the same, up to isomorphism, as quotient groups by normal subgroups. If $f:G\to G_1$ is onto, then $G_1\cong G/\ker f$, and the kernel is normal in $G$. Conversely, every normal subgroup gives a natural homomorphism onto a quotient group. This lesson turns the first isomorphism theorem into a classification principle for possible epimorphic images. :::definition[Homomorphic Image] Let $G$ and $G_1$ be groups. The group $G_1$ is called a homomorphic image of $G$ if there exists an epimorphism $f:G\to G_1$. ::: :::theorem[Characterization of Homomorphic Images] A group $G_1$ is a homomorphic image of $G$ if and only if there exists a normal subgroup $N\trianglelefteq G$ such that $$\begin{equation*} G/N\cong G_1. \end{equation*}$$ ::: :::proof Given that $G$ and $G_1$ are groups. To prove that $G_1$ is a homomorphic image of $G$ if and only if $G/N\cong G_1$ for some $N\trianglelefteq G$. First suppose that $G_1$ is a homomorphic image of $G$. Then there exists an epimorphism $f:G\to G_1$. Put $N=\ker f$. Since kernels are normal subgroups, $N\trianglelefteq G$. By the first isomorphism theorem, $$\begin{equation*} G/N=G/\ker f\cong G_1. \end{equation*}$$ Conversely, suppose there exists $N\trianglelefteq G$ such that $G/N\cong G_1$. Let $\pi:G\to G/N$ be the natural homomorphism. Then $\pi$ is onto. Let $\alpha:G/N\to G_1$ be an isomorphism. The composition $$\begin{equation*} \alpha\circ\pi:G\to G_1 \end{equation*}$$ is an epimorphism, because both $\pi$ and $\alpha$ are onto. Therefore $G_1$ is a homomorphic image of $G$. Hence the characterization is proved. $\square$ ::: A homomorphic image may be smaller or simpler than the original group, but some structure survives. If the original group is cyclic, the image is cyclic. If the original group is commutative, the image is commutative. These statements follow from onto-ness: every element in the image group comes from an element of the original group. A common mistake is to reverse these implications. A noncyclic group can have a cyclic quotient, and a noncommutative group can have a commutative quotient. :::theorem[Homomorphic Image of a Cyclic Group] Let $f:G\to G_1$ be an epimorphism. If $G$ is cyclic, then $G_1$ is cyclic. ::: :::proof Given that $f:G\to G_1$ is an epimorphism and $G$ is cyclic. To prove that $G_1$ is cyclic. Let $G=\langle a\rangle$. Since $f$ is onto, for every $y\in G_1$ there exists $x\in G$ such that $f(x)=y$. Since $G=\langle a\rangle$, there exists $n\in\mathbb{Z}$ such that $x=a^n$. Hence $$\begin{align*} y&=f(x)\\ &=f(a^n)\\ &=f(a)^n. \end{align*}$$ Thus every element of $G_1$ is a power of $f(a)$. Therefore $G_1=\langle f(a)\rangle$. Hence $G_1$ is cyclic. $\square$ ::: :::theorem[Homomorphic Image of a Commutative Group] Let $f:G\to G_1$ be an epimorphism. If $G$ is commutative, then $G_1$ is commutative. ::: :::proof Given that $f:G\to G_1$ is an epimorphism and $G$ is commutative. To prove that $G_1$ is commutative. Let $x,y\in G_1$. Since $f$ is onto, there exist $a,b\in G$ such that $x=f(a)$ and $y=f(b)$. Then $$\begin{align*} xy&=f(a)f(b)\\ &=f(ab)\\ &=f(ba) \quad [\because G \text{ is commutative}]\\ &=f(b)f(a)\\ &=yx. \end{align*}$$ Hence $G_1$ is commutative. $\square$ ::: :::example[Homomorphic Images of S Three] The homomorphic images of $S_3$ correspond to quotient groups $S_3/N$, where $N\trianglelefteq S_3$. The normal subgroups of $S_3$ are $$\begin{equation*} \{e\},\quad A_3,\quad S_3. \end{equation*}$$ Thus the homomorphic images, up to isomorphism, are $$\begin{equation*} S_3/\{e\}\cong S_3,\qquad S_3/A_3\cong\mathbb{Z}_2,\qquad S_3/S_3\cong\mathbb{Z}_1. \end{equation*}$$ So the only homomorphic images of $S_3$, up to isomorphism, are $S_3$, $\mathbb{Z}_2$, and the trivial group. ::: This preview shows how a normal subgroup determines a homomorphic image. Select one of the normal subgroups of $S_3$ and observe the quotient that it produces. The subgroup selected as the kernel is the part of $S_3$ collapsed to the identity in the image. Try moving from $\{e\}$ to $A_3$ to $S_3$ and watch the image become smaller. This is the classification principle: different normal kernels give the possible epimorphic images. :::scientific-preview[Normal Kernel Image Explorer] ::: The selected normal subgroup plays the role of a kernel. Enlarging the kernel collapses more of the original group, so the resulting homomorphic image becomes smaller. This is why listing normal subgroups is the correct first step when listing homomorphic images. This calculator tests the most common finite cyclic case. For cyclic groups, an epimorphism from $\mathbb Z_n$ onto $\mathbb Z_m$ exists exactly when $m$ divides $n$. Enter $n$ and $m$ and compare the target size with the kernel size. The output explains why a cyclic source can only map onto cyclic targets whose order divides the source order. :::calculator[Cyclic Epimorphism Checker] ::: The divisibility condition is the finite cyclic version of the normal-subgroup classification. A successful epimorphism has a kernel of the right size, and the quotient by that kernel has exactly the target order. :::solved-problem Show that there is no epimorphism from $\mathbb{Z}_4$ onto the Klein four-group $V$. ::: :::solution Let $V$ be the Klein four-group. If there were an epimorphism $f:\mathbb{Z}_4\to V$, then $V$ would be a homomorphic image of $\mathbb{Z}_4$. Since $\mathbb{Z}_4$ is cyclic, every homomorphic image of $\mathbb{Z}_4$ must be cyclic. But $V$ is not cyclic. Each nonidentity element of $V$ has order $2$, so no element generates all four elements of $V$. This contradiction shows that no epimorphism from $\mathbb{Z}_4$ onto $V$ exists. $\square$ ::: :::solved-problem Show that there is no epimorphism from the Klein four-group $V$ onto $\mathbb{Z}_4$. ::: :::solution Let $V$ be the Klein four-group. Suppose there exists an epimorphism $f:V\to\mathbb{Z}_4$. The group $\mathbb{Z}_4$ has an element of order $4$, namely $[1]$. Since $f$ is onto, there exists $x\in V$ such that $f(x)=[1]$. Then $o(f(x))=4$. For every homomorphism, the order of the image of an element divides the order of the element. Hence $4\mid o(x)$. This is impossible because every element of $V$ has order $1$ or $2$. Therefore no epimorphism from $V$ onto $\mathbb{Z}_4$ exists. $\square$ ::: :::tip To list homomorphic images of a group, list its normal subgroups first. To rule out an epimorphism, compare structural features that must survive under epimorphic images, such as cyclicity, commutativity, and possible element orders. ::: :::exercise 1. List the homomorphic images of a simple group $G$, up to isomorphism. 2. Prove that every homomorphic image of an abelian group is abelian. 3. Can a nonabelian group have an abelian homomorphic image? ::: :::answer 1. A simple group has only $\{e\}$ and $G$ as normal subgroups, so its homomorphic images are isomorphic to $G$ and to the trivial group. 2. This is exactly the homomorphic image of a commutative group theorem. 3. Yes. For example, $S_3/A_3\cong\mathbb{Z}_2$. ::: :::faq[Frequently Asked Questions] Q: Are homomorphic images exactly quotients? A: Up to isomorphism, yes. Every homomorphic image is a quotient by a normal subgroup. Q: Why must the subgroup be normal? A: Kernels are normal, and quotient groups require normal subgroups. Q: Can a quotient have more structure than the original group? A: It cannot gain properties that epimorphic images cannot support, such as cyclicity from a noncyclic source in reverse; structural implications must be checked carefully. ::: :::call-to-action[Continue to the Second Isomorphism Theorem] subtitle: Compare a subgroup with a quotient by a normal subgroup through the isomorphism $H/(H\cap K)\cong HK/K$. button: Study the Second Isomorphism Theorem | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/isomorphism-and-correspondence-theorems/second-isomorphism-theorem button-ghost: Review the First Isomorphism Theorem | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/isomorphism-and-correspondence-theorems/first-isomorphism-theorem :::