After studying kernels, we have seen that every kernel is a normal subgroup. The next statement goes in the reverse direction: every normal subgroup occurs as a kernel. The focus keyword natural homomorphism refers to the canonical map from a group onto a quotient group. If $H\trianglelefteq G$, the quotient group $G/H$ consists of cosets of $H$, and the natural homomorphism sends an element $a$ to the coset $aH$. This map is simple, but it is one of the main bridges between homomorphisms and quotient groups. In this lesson we prove that the map is a homomorphism and compute its kernel exactly. :::definition[Natural Homomorphism] Let $G$ be a group and let $H\trianglelefteq G$. The natural homomorphism from $G$ onto $G/H$ is the function $\pi:G\to G/H$ defined by $$\begin{equation*} \pi(a)=aH \end{equation*}$$ for every $a\in G$. ::: The quotient $G/H$ is a group only when $H$ is normal in $G$. Without normality, multiplication of cosets may not be well-defined. The natural homomorphism therefore begins with $H\trianglelefteq G$, not merely $H\leq G$. A common classroom error is to define $a\mapsto aH$ for any subgroup and too quickly call the codomain a quotient group. The formula makes sense as a set-valued assignment to cosets, but the group homomorphism statement needs the quotient group operation. :::theorem[Natural Homomorphism Theorem] Let $H\trianglelefteq G$ and define $\pi:G\to G/H$ by $\pi(a)=aH$. Then $\pi$ is an epimorphism and $$\begin{equation*} \ker \pi=H. \end{equation*}$$ ::: :::proof Given that $H\trianglelefteq G$ and $\pi:G\to G/H$ is defined by $\pi(a)=aH$. To prove that $\pi$ is an epimorphism and $\ker \pi=H$. Let $a,b\in G$. Then $$\begin{align*} \pi(ab)&=(ab)H,\\ \pi(a)\pi(b)&=(aH)(bH)\\ &=(ab)H. \end{align*}$$ Therefore $\pi(ab)=\pi(a)\pi(b)$, so $\pi$ is a homomorphism. To prove that $\pi$ is onto, let $aH\in G/H$. Then $aH=\pi(a)$, so every element of $G/H$ lies in the image of $\pi$. Now the identity element of $G/H$ is $H=eH$. Hence $$\begin{align*} \ker \pi&=\{a\in G\mid \pi(a)=H\}\\ &=\{a\in G\mid aH=H\}. \end{align*}$$ For $a\in G$, $aH=H$ if and only if $a\in H$. Thus $$\begin{equation*} \ker \pi=H. \end{equation*}$$ Hence $\pi$ is an epimorphism and $\ker \pi=H$. $\square$ ::: :::example[Natural Homomorphism from the Integers] Let $n$ be a positive integer. Since $n\mathbb{Z}\trianglelefteq\mathbb{Z}$, define $\pi:\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ by $$\begin{equation*} \pi(a)=a+n\mathbb{Z}. \end{equation*}$$ For $a,b\in\mathbb{Z}$, $$\begin{align*} \pi(a+b)&=(a+b)+n\mathbb{Z},\\ \pi(a)+\pi(b)&=(a+n\mathbb{Z})+(b+n\mathbb{Z})\\ &=(a+b)+n\mathbb{Z}. \end{align*}$$ Thus $\pi$ is a homomorphism. Its kernel is $$\begin{align*} \ker \pi&=\{a\in\mathbb{Z}\mid a+n\mathbb{Z}=n\mathbb{Z}\}\\ &=n\mathbb{Z}. \end{align*}$$ ::: :::example[Natural Homomorphism from S Three] Let $H=A_3=\{e,(123),(132)\}$ in $S_3$. Since $A_3\trianglelefteq S_3$, define $\pi:S_3\to S_3/A_3$ by $\pi(\sigma)=\sigma A_3$. Then $\pi$ is a natural homomorphism. Its kernel is $$\begin{equation*} \ker \pi=A_3. \end{equation*}$$ This example shows concretely that the normal subgroup being divided out is exactly the part collapsed to the identity coset. ::: :::solved-problem Let $G$ be a group and let $H\trianglelefteq G$. Prove that the natural homomorphism $\pi:G\to G/H$ identifies two elements $a,b\in G$ precisely when $aH=bH$. ::: :::solution Let $a,b\in G$, let $H\trianglelefteq G$, and let $\pi:G\to G/H$ be defined by $\pi(x)=xH$. To prove that $\pi(a)=\pi(b)$ if and only if $aH=bH$. Let $a,b\in G$. By the definition of $\pi$, $$\begin{equation*} \pi(a)=aH \end{equation*}$$ and $$\begin{equation*} \pi(b)=bH. \end{equation*}$$ Therefore $$\begin{align*} \pi(a)=\pi(b)&\Longleftrightarrow aH=bH. \end{align*}$$ Hence the natural homomorphism identifies exactly the elements lying in the same coset of $H$. $\square$ ::: :::remark The natural homomorphism gives the conceptual reason quotient groups are linked to normal subgroups. The subgroup $H$ becomes the identity element of $G/H$, and all elements of the same coset have the same image. This is why quotient groups are often described as groups obtained by collapsing $H$ to the identity. ::: :::exercise 1. Let $4\mathbb{Z}\trianglelefteq\mathbb{Z}$. Write the natural homomorphism from $\mathbb{Z}$ onto $\mathbb{Z}/4\mathbb{Z}$ and find its kernel. 2. Let $H\trianglelefteq G$. Is the natural homomorphism $\pi:G\to G/H$ always onto? 3. If the natural homomorphism $\pi:G\to G/H$ has identity-only kernel, what is $H$? ::: :::answer 1. The map is $\pi(a)=a+4\mathbb{Z}$, and its kernel is $4\mathbb{Z}$. 2. Yes. Every coset $aH$ is equal to $\pi(a)$. 3. Since $\ker \pi=H$, a identity-only kernel means $H=\{e\}$. ::: :::faq[Frequently Asked Questions] Q: Why is the kernel of the natural homomorphism equal to $H$? A: Because the identity in $G/H$ is the coset $H$, and $aH=H$ exactly when $a\in H$. Q: Is the natural homomorphism always one-to-one? A: No. It is one-to-one only when $H=\{e\}$. Q: Why is it called natural? A: The definition uses the quotient construction itself; no arbitrary choices are needed. ::: :::call-to-action[Continue to Isomorphism Tests] subtitle: Learn how bijective homomorphisms prove structural sameness and how preserved properties disprove isomorphism. button: Study Isomorphism Tests | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/homomorphisms-of-groups/isomorphism-tests-and-examples button-ghost: Review Quotient Groups | /abstract-algebra/subgroups-and-normal-subgroups/quotient-groups/quotient-groups :::