After proving the basic properties of homomorphisms, the next natural question is which elements disappear under a homomorphism. The focus keyword kernel of a homomorphism refers to the set of domain elements whose image is the identity element of the codomain. The kernel measures the amount of collapse caused by the map. If the kernel is only the identity, the homomorphism keeps different elements separated. If the kernel contains more than the identity, several elements of the domain become indistinguishable in the codomain. In this lesson we define the kernel, prove that it is a normal subgroup, and use it to test one-to-one homomorphisms. :::definition[Kernel of a Homomorphism] Let $f:G\to G_1$ be a homomorphism of groups, and let $e_1$ be the identity element of $G_1$. The kernel of the homomorphism $f$ is the set $$\begin{equation*} \ker f=\{a\in G\mid f(a)=e_1\}. \end{equation*}$$ Equivalently, $\ker f=f^{-1}(\{e_1\})$. ::: Finding a kernel is usually an equation-solving task. First identify the identity element of the codomain, not the identity of the domain. Then solve $f(a)=e_1$ inside the domain. Students often write the identity of the domain by habit; this is harmless only when both groups use the same notation. The kernel of a homomorphism is not a random subset. It is always a normal subgroup, and this fact is one reason kernels are central in quotient group theory. :::theorem[Kernel Is a Normal Subgroup] Let $f:G\to G_1$ be a homomorphism. Then $$\begin{equation*} \ker f\trianglelefteq G. \end{equation*}$$ ::: :::proof Given that $f:G\to G_1$ is a homomorphism. To prove that $\ker f\trianglelefteq G$. Let $e$ be the identity element of $G$, and let $e_1$ be the identity element of $G_1$. First, $f(e)=e_1$, so $e\in\ker f$. Therefore $\ker f\neq\varnothing$. Let $a,b\in\ker f$. Then $f(a)=e_1$ and $f(b)=e_1$. Hence $$\begin{align*} f(ab^{-1})&=f(a)f(b^{-1})\\ &=f(a)f(b)^{-1}\\ &=e_1e_1^{-1}\\ &=e_1. \end{align*}$$ Thus $ab^{-1}\in\ker f$. Therefore $\ker f\leq G$. To prove normality, let $x\in\ker f$ and $g\in G$. Then $$\begin{align*} f(gxg^{-1})&=f(g)f(x)f(g^{-1})\\ &=f(g)e_1f(g)^{-1}\\ &=e_1. \end{align*}$$ Thus $gxg^{-1}\in\ker f$ for all $g\in G$ and all $x\in\ker f$. Hence $\ker f\trianglelefteq G$. $\square$ ::: :::theorem[Kernel Test for One-to-One Homomorphisms] Let $f:G\to G_1$ be a homomorphism. Then $f$ is one-to-one if and only if $$\begin{equation*} \ker f=\{e\}, \end{equation*}$$ where $e$ is the identity element of $G$. ::: :::proof Given that $f:G\to G_1$ is a homomorphism. To prove that $f$ is one-to-one if and only if $\ker f=\{e\}$. First suppose that $f$ is one-to-one. Since $f(e)=e_1$, we have $e\in\ker f$. If $a\in\ker f$, then $$\begin{equation*} f(a)=e_1=f(e). \end{equation*}$$ Since $f$ is one-to-one, $a=e$. Therefore $\ker f=\{e\}$. Conversely, suppose $\ker f=\{e\}$. Let $a,b\in G$ and suppose $f(a)=f(b)$. Then $$\begin{align*} f(ab^{-1})&=f(a)f(b^{-1})\\ &=f(a)f(b)^{-1}\\ &=e_1. \end{align*}$$ Thus $ab^{-1}\in\ker f$. Since $\ker f=\{e\}$, $ab^{-1}=e$, so $a=b$. Hence $f$ is one-to-one if and only if $\ker f=\{e\}$. $\square$ ::: :::example[Kernel of the Natural Map from Integers] Let $n$ be a positive integer and define $f:(\mathbb{Z},+)\to(\mathbb{Z}_n,+_n)$ by $f(a)=[a]$. The identity element of $\mathbb{Z}_n$ is $[0]$. Therefore $$\begin{align*} \ker f&=\{a\in\mathbb{Z}\mid [a]=[0]\}\\ &=\{a\in\mathbb{Z}\mid n\mid a\}\\ &=n\mathbb{Z}. \end{align*}$$ Thus the kernel is the subgroup of all multiples of $n$. ::: :::example[Kernel of the Absolute Value Homomorphism] Let $f:(\mathbb{R}\setminus\{0\},\cdot)\to(\mathbb{R}^{+},\cdot)$ be defined by $f(a)=|a|$. The identity element of $\mathbb{R}^{+}$ is $1$. Hence $$\begin{align*} \ker f&=\{a\in\mathbb{R}\setminus\{0\}\mid |a|=1\}\\ &=\{1,-1\}. \end{align*}$$ Since the kernel is not $\{1\}$, this homomorphism is not one-to-one. ::: :::solved-problem Let $f:G\to G_1$ be a homomorphism. Prove that for $a,b\in G$, $f(a)=f(b)$ if and only if $ab^{-1}\in\ker f$. ::: :::solution Let $a,b\in G$, and let $f:G\to G_1$ be a homomorphism. To prove that $f(a)=f(b)$ if and only if $ab^{-1}\in\ker f$. First suppose $f(a)=f(b)$. Then $$\begin{align*} f(ab^{-1})&=f(a)f(b^{-1})\\ &=f(a)f(b)^{-1}\\ &=e_1. \end{align*}$$ Therefore $ab^{-1}\in\ker f$. Conversely, suppose $ab^{-1}\in\ker f$. Then $f(ab^{-1})=e_1$, so $$\begin{align*} f(a)f(b)^{-1}&=e_1. \end{align*}$$ Multiplying by $f(b)$ on the right gives $f(a)=f(b)$. Hence $f(a)=f(b)$ if and only if $ab^{-1}\in\ker f$. $\square$ ::: :::tip The kernel of a homomorphism is always computed in the domain, even though the defining equation uses the identity of the codomain. This distinction prevents a common error: writing $\ker f=\{e_1\}$. The element $e_1$ may not even belong to the domain. ::: :::exercise 1. Find the kernel of $f:(\mathbb{Z},+)\to(\mathbb{Z},+)$ defined by $f(a)=3a$. 2. Find the kernel of $g:(\mathbb{R}^{+},\cdot)\to(\mathbb{R}^{+},\cdot)$ defined by $g(a)=a^2$. 3. If $\ker f\neq\{e\}$, can $f$ be one-to-one? ::: :::answer 1. The identity of the codomain is $0$, so $3a=0$ gives $a=0$. Hence $\ker f=\{0\}$. 2. The identity of the codomain is $1$, so $a^2=1$ with $a\in\mathbb{R}^{+}$ gives $a=1$. Hence $\ker g=\{1\}$. 3. No. A homomorphism is one-to-one if and only if its kernel is $\{e\}$. ::: :::faq[Frequently Asked Questions] Q: Is every normal subgroup a kernel? A: Yes, every normal subgroup is the kernel of the natural homomorphism onto the quotient group. Q: Can a kernel be empty? A: No. Since $f(e)=e_1$, the identity element of the domain always lies in the kernel. Q: Does a identity-only kernel mean the homomorphism is onto? A: No. A identity-only kernel means one-to-one, not onto. ::: :::call-to-action[Continue to Special Homomorphisms] subtitle: Distinguish epimorphisms, monomorphisms, isomorphisms, and automorphisms using onto and one-to-one conditions. button: Study Special Homomorphisms | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/homomorphisms-of-groups/epimorphism-monomorphism-isomorphism-automorphism button-ghost: Review Quotient Groups | /abstract-algebra/subgroups-and-normal-subgroups/quotient-groups/quotient-groups :::