# The Kernel of a Group Homomorphism

Recall from the Group Homomorphisms page that if $(G_1, *_1)$ and $(G_2, *_2)$ are groups then a homomorphism between these groups is a function $f : G_1 \to G_2$ such that for all $x, y \in G_1$ we have that:

(1)On the Basic Theorems Regarding Group Homomorphisms page we looked at some important properties of group homomorphisms. For example, if $f : G_1 \to G_2$ is a homomorphism and if $e_1$ is the identity of $G_1$ then the identity of $G_2$ is $e_2 = f(e_1)$. Furthermore, for each $x \in G_1$ we have that $f(x^{-1}) = [f(x)]^{-1}$.

We also noted that if $H$ is a subgroup of $G_1$ then $f(H)$ is a subgroup of $G_2$ and similarly, if $H$ is a subgroup of $G_2$ then $f^{-1} (H)$ is a subgroup of $G_1$.

Lastly, we noted that if $H$ is a normal subgroup of $G_2$ then $f^{-1}(H)$ is a normal subgroup of $G_1$. Note that for any group $G_2$ that if $e_2$ is the identity in $G_2$ then $H= \{ e_2 \}$ is a normal subgroup of $G_2$ (albiet rather trivial), so provided that a group homomorphism exists between $G_1$ and $G_2$, we can construct a normal subgroup of the domain $G_1$ to be $f^{-1} (\{ e_2 \})$. This subgroup is very important and given a special name.

Definition: Let $(G_1, *_1)$ and $(G_2, *_2)$ be two groups and let $f : G_1 \to G_2$ be a group homomorphism. Let $e_2$ be the identity of $G_2$. Then the Kernel of $f$ is defined as the set/subgroup (of $G_1$) denoted $\mathrm{ker} (f) = f^{-1} ( \{ e_2 \} ) = \{ x \in G_1 : f(x) = e_2 \}$. |

*Note that $\mathrm{ker} (f) \neq \emptyset$ since we know that if $e_1$ is the identity of $G_1$ 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.

Thereom 1: Let $(G_1, *_1)$ and $(G_2, *_2)$ be groups and let $f : G_1 \to G_2$ be a group homomorphism. Let $e_1$ be the identity of $G_1$ and let $e_2$ be the identity of $G_2$. If $\mathrm{ker} (f) = \{ e_1 \}$ then $f$ is injective. |

**Proof:**Suppose that $\mathrm{ker} (f) = \{ e_1 \}$. Let $x, y \in G_1$ and suppose that $f(x) = f(y)$. Then $f(x) *_2 [f(y)]^{-1} = e_2$. But then

- So $x *_1 y^{-1} \in \mathrm{ker} (f)$. But $\mathrm{ker} (f) = \{ e_1 \}$, so $x *_1 y^{-1} = e_1$, i.e., $x = y$. So $f$ is injective. $\blacksquare$

Corollary 2: Let $(G_1, *_1)$ and $(G_2, *_2)$ be groups and let $f : G_1 \to G_2$ be a group homomorphism. Let $e_1$ be the identity of $G_1$ and let $e_2$ be the identity of $G_2$. 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 theorem 1, $f$ is injective.

- Otherwise $\mathrm{ker} (f) = G$ which means that for all $x \in G$, $f(x) = e_2$. Thus $f(G) = \{ e_2 \}$. $\blacksquare$