Group homomorphism/Kernel/Injectivity criterion/Introduction/Section

Definition

Let ${}G$ and ${}H$ be groups, and let

$\varphi \colon G\longrightarrow H$

be a group homomorphism. Then the preimage of the neutral element is called the kernel of ${}\varphi$, denoted by

${}\operatorname {kern} \varphi =\varphi ^{-1}(e_{H})={\left\{g\in G\mid \varphi (g)=e_{H}\right\}}\,.$

Lemma

Let ${}G$ and ${}H$ be groups, and let

$\varphi \colon G\longrightarrow H$

be a group homomorphism. Then the kernel of ${}\varphi$ is a subgroup

of ${}G$.

Proof

Because of ${}\varphi (e_{G})=e_{H}$, we have ${}e_{G}\in \operatorname {kern} \varphi$. Let ${}g,g'\in \operatorname {kern} \varphi$. Then

${}\varphi (gg')=\varphi (g)\varphi (g')=e_{H}e_{H}=e_{H}\,;$

therefore, also ${}gg'\in \operatorname {kern} \varphi$. Hence, the kernel is a submonoid. Now, let ${}g\in \operatorname {kern} \varphi$, and consider the inverse element ${}g^{-1}$. Due to fact, we have

${}\varphi {\left(g^{-1}\right)}=(\varphi (g))^{-1}=e_{H}^{-1}=e_{H}\,;$

Hence, ${}g^{-1}\in \operatorname {kern} \varphi$.

$\Box$

Lemma

Let ${}G$ and ${}H$ be groups. A group homomorphism ${}\varphi \colon G\rightarrow H$ is injective if and only if the kernel

of ${}\varphi$ is trivial.

Proof

If ${}\varphi$ is injective, then every element ${}h\in H$ is hit by at most one element from ${}G$. As ${}e_{G}$ is sent to ${}e_{H}$, no further element can be sent to ${}e_{H}$. Therefore, ${}\operatorname {kern} \varphi =\{e_{G}\}$. Now assume that this holds. Let ${}g,{\tilde {g}}\in G$ be elements mapping to ${}h\in H$. Then

${}\varphi {\left(g{\tilde {g}}^{-1}\right)}=\varphi (g)\varphi ({\tilde {g}})^{-1}=hh^{-1}=e_{H}\,;$

hence, ${}g{\tilde {g}}^{-1}\in \operatorname {kern} \varphi$, and so ${}g{\tilde {g}}^{-1}=e_{G}$ by the condition. Therefore, ${}g={\tilde {g}}$.

$\Box$