Having established factorization through quotients, we now choose the most important possible quotient: the quotient by the kernel. The focus keyword first isomorphism theorem says that $G/\ker f$ has exactly the same structure as the image $f(G)$. The kernel is precisely the part of $G$ collapsed to the identity, so quotienting by it removes exactly the ambiguity created by $f$. Students often memorize the formula but miss the idea: the quotient identifies exactly those elements that have the same image. This theorem is the central bridge from homomorphisms to quotient groups. :::theorem[First Isomorphism Theorem] Let $f:G\to G_1$ be a homomorphism. Then $$\begin{equation*} G/\ker f\cong f(G). \end{equation*}$$ If $f$ is onto, then $$\begin{equation*} G/\ker f\cong G_1. \end{equation*}$$ ::: :::proof Given that $f:G\to G_1$ is a homomorphism. To prove that $G/\ker f\cong f(G)$, and that $G/\ker f\cong G_1$ when $f$ is onto. Let $N=\ker f$. Since kernels are normal subgroups, $N\trianglelefteq G$. Define $\Phi:G/N\to f(G)$ by $$\begin{equation*} \Phi(aN)=f(a). \end{equation*}$$ [1] To prove that $\Phi$ is well-defined. Suppose $aN=bN$. Then $b^{-1}a\in N=\ker f$. Therefore $f(b^{-1}a)=e_1$, and $$\begin{align*} f(b)^{-1}f(a)&=e_1,\\ f(a)&=f(b). \end{align*}$$ Thus $\Phi(aN)=\Phi(bN)$. [2] To prove that $\Phi$ is a homomorphism. Let $aN,bN\in G/N$. Then $$\begin{align*} \Phi((aN)(bN))&=\Phi(abN)\\ &=f(ab)\\ &=f(a)f(b)\\ &=\Phi(aN)\Phi(bN). \end{align*}$$ [3] To prove that $\Phi$ is onto. Let $y\in f(G)$. Then $y=f(a)$ for some $a\in G$. Hence $y=\Phi(aN)$. [4] To prove that $\Phi$ is one-to-one. If $\Phi(aN)=e_1$, then $f(a)=e_1$, so $a\in\ker f=N$. Therefore $aN=N$. Thus $\ker\Phi=\{N\}$, and $\Phi$ is one-to-one. Since $\Phi$ is a one-to-one and onto homomorphism, it is an isomorphism. Hence $$\begin{equation*} G/\ker f\cong f(G). \end{equation*}$$ If $f$ is onto, then $f(G)=G_1$, so $G/\ker f\cong G_1$. $\square$ ::: The first isomorphism theorem measures a homomorphism in two parts: what it kills and what it reaches. The kernel tells us which elements become identity after applying $f$, while the image tells us which elements of the codomain are actually reached. Quotienting by the kernel repairs the loss of injectivity. The result is not always the whole codomain; it is the image. Only when $f$ is onto may we replace $f(G)$ by $G_1$. :::example[The Natural Map from Z to Z Three] Let $f:(\mathbb{Z},+)\to(\mathbb{Z}_3,+_3)$ be defined by $f(n)=[n]_3$. The kernel is $$\begin{equation*} \ker f=(3). \end{equation*}$$ The map is onto because every element of $\mathbb{Z}_3$ is $[n]_3$ for some $n\in\mathbb{Z}$. By the first isomorphism theorem, $$\begin{equation*} \mathbb{Z}/(3)\cong\mathbb{Z}_3. \end{equation*}$$ This shows that the usual residue class group is exactly the quotient of $\mathbb{Z}$ by the subgroup of multiples of $3$. ::: :::example[Cyclic Groups as Quotients of Z] Let $G=\langle a\rangle$ be a cyclic group and define $f:\mathbb{Z}\to G$ by $f(n)=a^n$. Then $$\begin{equation*} f(m+n)=a^{m+n}=a^ma^n=f(m)f(n), \end{equation*}$$ so $f$ is a homomorphism. It is onto because $G=\langle a\rangle$. If $o(a)=n$, then $\ker f=(n)$, and the first isomorphism theorem gives $$\begin{equation*} \mathbb{Z}/(n)\cong G. \end{equation*}$$ If $a$ has infinite order, then $\ker f=\{0\}$ and $\mathbb{Z}\cong G$. ::: This preview models a homomorphism $\phi:\mathbb{Z}_n\to\mathbb{Z}_m$ by the rule $\phi([x])=[kx]_m$. Choose $n$, $m$, and the multiplier $k$, then observe how the elements of $\mathbb{Z}_n$ group into fibers with the same image. The theorem says that each fiber is a coset of the kernel, so collapsing the kernel leaves exactly one representative for each image value. Start with $n=12$, $m=8$, and $k=2$, then try changing $k$ to see how the image and kernel change together. :::scientific-preview[Kernel Fiber Explorer] ::: The equality of sizes is not the theorem by itself, but it reveals the mechanism. The kernel identifies exactly the elements that should be treated as equivalent, and the quotient has one class for each image value. :::solved-problem Let $f:G\to G_1$ be an onto homomorphism. Prove that $G/\ker f\cong G_1$. ::: :::solution Let $f:G\to G_1$ be an onto homomorphism. By the first isomorphism theorem, $$\begin{equation*} G/\ker f\cong f(G). \end{equation*}$$ Since $f$ is onto, every element of $G_1$ lies in $f(G)$. Hence $$\begin{equation*} f(G)=G_1. \end{equation*}$$ Substituting this into the isomorphism gives $$\begin{equation*} G/\ker f\cong G_1. \end{equation*}$$ $\square$ ::: :::solved-problem Let $f:S_3\to\{1,-1\}$ be the sign homomorphism. Use the first isomorphism theorem to identify $S_3/\ker f$. ::: :::solution Let $f:S_3\to\{1,-1\}$ be the sign homomorphism. The kernel of $f$ consists of the even permutations, so $$\begin{equation*} \ker f=A_3. \end{equation*}$$ The map is onto because both $1$ and $-1$ occur as signs of permutations in $S_3$. By the first isomorphism theorem, $$\begin{equation*} S_3/A_3\cong\{1,-1\}. \end{equation*}$$ Since $\{1,-1\}$ under multiplication is cyclic of order $2$, we also have $$\begin{equation*} S_3/A_3\cong\mathbb{Z}_2. \end{equation*}$$ $\square$ ::: :::tip When applying the first isomorphism theorem, compute the kernel and the image separately. Do not assume the image is the full codomain unless the homomorphism is known to be onto. ::: :::exercise 1. Let $f:\mathbb{Z}\to\mathbb{Z}_6$ be given by $f(n)=[n]_6$. Use the first isomorphism theorem. 2. Let $f:\mathbb{R}\to\mathbb{R}^{+}$ be $f(x)=e^x$. Identify $\ker f$ and the quotient isomorphic to the image. 3. If $f:G\to G_1$ is one-to-one, what does the first isomorphism theorem say? ::: :::answer 1. Since $\ker f=(6)$ and $f$ is onto, $\mathbb{Z}/(6)\cong\mathbb{Z}_6$. 2. The kernel is $\{0\}$ and the image is $\mathbb{R}^{+}$, so $\mathbb{R}/\{0\}\cong\mathbb{R}^{+}$. 3. If $f$ is one-to-one, then $\ker f=\{e\}$, so $G/\{e\}\cong f(G)$, hence $G\cong f(G)$. ::: :::faq[Frequently Asked Questions] Q: Why is the quotient by the kernel used? A: The kernel is exactly the part of the domain collapsed to the identity, so quotienting by it removes precisely that collapse. Q: Is the theorem about the codomain or the image? A: It is about the image. The codomain appears only when the homomorphism is onto. Q: Does the theorem prove every image is a quotient? A: Yes. Every homomorphic image is isomorphic to a quotient by the kernel. ::: :::call-to-action[Continue to Homomorphic Images] subtitle: Use the first isomorphism theorem to characterize all homomorphic images of a group through its normal subgroups. button: Study Homomorphic Images | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/isomorphism-and-correspondence-theorems/homomorphic-images-and-normal-subgroups button-ghost: Review Factorization | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/isomorphism-and-correspondence-theorems/factorization-through-quotients :::