Kernel of a homomorphism

The kernel is always a subgroup of \(G\): it contains \(e_G\), and if \(\varphi(a)=\varphi(b)=e_H\), then \(\varphi(ab^{-1})=\varphi(a)\varphi(b)^{-1}=e_H\). It is also closed under conjugation, a property that will later make it the central example of a normal subgroup.

The kernel detects injectivity: \(\varphi\) is injective if and only if \(\ker\varphi=\{e_G\}\). Indeed, \(\varphi(a)=\varphi(b)\) is equivalent to \(\varphi(ab^{-1})=e_H\), that is, to \(ab^{-1}\) lying in the kernel. So checking that a homomorphism is injective reduces to a single equation.

More precisely, two elements have the same image exactly when they differ by an element of the kernel, so each value of the image has as many preimages as \(\ker\varphi\) has elements. Hence, if \(G\) is finite, \(|G|=|\ker\varphi|\cdot|\operatorname{im}\varphi|\). For example, \(x\mapsto x\bmod 4\) from \(\mathbb{Z}_{12}\) to \(\mathbb{Z}_4\) has kernel \(\{0,4,8\}\) and an image with \(4\) elements.

A typical mistake is to confuse the kernel with the image, or to look for the identity of the domain instead of the codomain: for \(x\mapsto e^x\) the target identity is \(1\), yet the kernel is \(\{0\}\). The kernel is the piece that, together with cosets, will let us build the quotient group and state the first isomorphism theorem.