The Kernel of a Group Homomorphism

# The Kernel of a Group Homomorphism

 Definition: Let $(G, \cdot)$ and $(H, *)$ be two groups and let $f : G \to H$ be a group homomorphism. Let $e_2$ be the identity of $H$. Then the Kernel of $f$ is defined as the subgroup (of $G$) denoted $\mathrm{ker} (f) = f^{-1} ( \{ e_2 \} ) = \{ x \in G : f(x) = e_2 \}$.

Note that $\mathrm{ker} (f) \neq \emptyset$ since we know that if $e_1$ is the identity of $G$ then $f(e_1) = e_2$, i.e., $e_1 \in \mathrm{ker} (f)$.

The kernel of a group homomorphism has many nice attributes - some of which we acknowledge below.

 Proposition 1: Let $(G, \cdot)$ and $(H, *)$ be groups and let $f : G \to H$ be a group homomorphism. Then $f$ is injective if and only if $\ker(f) = \{ e_1 \}$.
• Proof: $\Rightarrow$ Suppose that $f$ is injective. Let $x \in \ker (f)$. Then $f(x) = e_2$. But since $f$ is a homomorphism we have that $f(e_1) = e_2$. Since $f$ is injective we must have that $x = e_1$ and so $\ker (f) = \{ e_1 \}$.
• $\Leftarrow$ Suppose that $\ker (f) = \{ e_1 \}$. Let $x, y \in G$ and suppose that $f(x) = f(y)$. Then $f(x) * [f(y)]^{-1} = e_2$. Since $f$ is a homomorphism we have that $f(x \cdot y^{-1}) = e_2$. So $x \cdot y^{-1} \in \ker (f)$, so $x \cdot y^{-1} = e_1$, i.e., $x = y$. So $f$ is injective. $\blacksquare$
 Corollary 2: Let $(G, \cdot)$ and $(H, *)$ be groups and let $f : G \to H$ be a group homomorphism. Let $e_1$ be the identity of $G$ and let $e_2$ be the identity of $H$. If $G$ is a simple group then either $f(G) = \{ e_2 \}$ or $f$ is injective.
• Proof: Since $G$ is a simple group and the kernel of $f$ is a subgroup of $G$ that must be normal, then $\mathrm{ker} (f) = \{ e_1 \}$ or $\mathrm{ker} (f) = G$ (since these are the only two normal subgroups of $G$).
• If $\mathrm{ker} (f) = \{ e_1 \}$ then by Proposition 1, $f$ is injective.
• Otherwise $\mathrm{ker} (f) = G$ which means that for all $x \in G$, $f(x) = e_2$. Thus $f(G) = \{ e_2 \}$. $\blacksquare$