The Kernel of a Group Homo. is a Normal Subgroup of the Domain
The Kernel of a Group Homomorphism is a Normal Subgroup of the Domain
| Lemma 1: Let $(G, *_1)$ and $(H, *_2)$ be groups with identities $1_G$ and $1_H$ respectively and let $\phi : G \to H$ be a homomorphism between them. Then $\ker (\phi)$ is a subgroup of domain, $G$. |
- Proof: By definition, $\ker (\phi) \subseteq G$. To show that $\ker (\phi)$ is a subgroup of $G$, all that is necessary to show is that $\ker (\phi)$ is closed under $*_1$ and such that for all $a \in \ker (\phi)$ we have that $a^{-1} \in \ker (\phi)$.
- Let $a, b \in \ker (\phi)$. Then $\phi (a) = 1_H$ and $\phi (b) = 1_H$. So:
\begin{align} \quad \phi (a *_1 b) = \phi(a) *_2 \phi(b) = 1_H * 1_H = 1_H \end{align}
- Therefore $(a *_1 b) \in \ker (\phi)$ so $\ker (\phi)$ is closed under $*_1$.
- Now let $a \in \ker (\phi)$. Then $\phi (a) = 1_H$. So:
\begin{align} \quad 1_H = \phi (1_G) = \phi(a *_1 a^{-1}) = \phi(a) *_2 \phi (a^{-1}) = 1_H * \phi(a^{-1}) = \phi(a^{-1}) \end{align}
- So $a^{-1} \in \ker (\phi)$. Hence $(\ker (\phi), *_1)$ is a subgroup of $(G, *_1)$. $\blacksquare$
| Theorem 2: Let $(G, *_1)$ and $(H, *_2)$ be groups with identities $1_G$ and $1_H$ respectively and let $\phi : G \to H$ be a homomorphism between them. Then $\ker (\phi)$ is a normal subgroup of domain, $G$. |
- Proof: From Lemma 1, $\ker (\phi)$ is a subgroup of $G$. All that remains to show is that this subgroup is normal. Let $a \in \ker (\phi)$ and let $g \in G$. Then:
\begin{align} \quad \phi(g *_1 a *_1 g^{-1}) = \phi (g) *_2 \phi(a) *_2 \phi (g^{-1}) = \phi (g) *_2 1_H *_2 \phi (g^{-1}) = \phi (g) *_2 \phi(g^{-1}) = \phi (g *_1 g^{-1}) = \phi(1_G) = 1_H \end{align}
- So $g *_1 a *_1 g^{-1} \in \ker (\phi)$ which implies from the Criteria for a Subgroup to be Normal page that $\ker (\phi)$ is a normal subgroup of $G$. $\blacksquare$