Math and science::Algebra::Aluffi

Groups. Kernel.

Kernel of a group homomorphism

Let $φ : G \to G^{'}$ be a group homomorphism from group $G$ to group $G^{'}$ . The kernel of $φ$ is the set of elements of $G$ that map to $e_{G^{'}}$ through $φ$ .

The kernel of $φ$ is denoted as $\ker φ$ .

In other words, $φ$ is the preimage of ${e_{G^{'}}}$ through $φ$ .

\ker φ = φ^{- 1} ({e_{G^{'}}})

From kernel to (equal) cosets

For the below discussion, it's worth remembering that kernels form subgroups, and these subgroups are normal.

Let $φ : G \to D$ be a group homomorphism. Consider the case if $\ker φ$ has 2 elements, $(e_{G}, h_{1})$ . The fact that $φ$ is a group homomorphism introduces some strong requirements on other elements of $G$ . Let $a \in G$ be an element not in $\ker (φ)$ . Then consider:

\begin{aligned} a \cdot e_{G} & = a \\ a \cdot h_{1} & = x \\ h_{1} \cdot a & = ? \end{aligned}

Note that $a h_{1} = h_{1} a$ . This commutativity is required so that $e_{D} x = x e_{D}$ in the group $D$ .

In the group $G$ , $e_{G}$ and $h_{1}$ are distinct elements, and so $a$ and $x$ must also be distinct elements, by the nature of groups. As $D$ is also a group and $φ$ a group homomorphism, $a$ and $x$ must map through $φ$ to the same element in $D$ . This is required to satisfy the requirement of group homomorphisms:

\forall w, v \in G, φ (w \cdot_{G} v) = φ (w) \cdot_{D} φ (v)

This means that for any $d \in D$ , the preimage of $d$ through $φ$ , $im φ ({d})$ , must have the same number of elements as $\ker φ$ ! We can use these preimages to form the blocks in a partition and the resulting equivalence relation is exactly the one corresponding to the quotient $G / \ker (φ)$ .

Below is a visualization of this idea.

Source

Aluffi p80