The kernel of a group homomorphism is a subgroup

The Mathematician — Wed, 01 Mar 2023 13:00:43 +0000

Let $G$ and $H$ be groups and let $\phi: G \longrightarrow H$ be a homomorphism. Then $\ker(\phi)$ is a subgroup of $G$ .

Proof. In order to show that

\ker(\phi)

is a subgroup of

H

, it must hold the subgroup criterion. For this proof, let

1_G

and

1_H

be the identities of

G

and

H

, respectively. Firstly, we know that

\ker(\phi)

is nonempty since it contains the identity element

1_G \in \ker(\phi)

. Secondly, let

x,y \in \ker(\phi)

. To satisfy the subgroup criterion, we only need to show that

xy^{-1} \in \ker(\phi)

. Since

\phi(x) = \phi(y) = 1_H

, we get the following:

\begin{align*} \phi(xy^{-1}) &= \phi(x)\phi(y^{-1}) \\ &= \phi(x)\phi(y)^{-1} \\ &= 1_H 1_H^{-1} \\ &= 1_H. \end{align*}

Therefore, this means that

xy^{-1} \in \ker(\phi)

. So

\ker(\phi)

is a subgroup of

G

by the subgroup criterion.

The post The kernel of a group homomorphism is a subgroup appeared first on Epsilonify.

kernel of a group homomorphism is a subgroup Archives - Epsilonify

The kernel of a group homomorphism is a subgroup