The image of a group homomorphism is a subgroup

The Mathematician — Sun, 05 Mar 2023 13:00:08 +0000

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

Proof. In order to show that

\text{im}(\phi)

is a subgroup of

H

, it must hold the subgroup criterion. Firstly, we know that

\text{im}(\phi)

is nonempty since

\phi(1_G) = 1_H

, where

1_G

and

1_H

are identities of

G

and

H

, respectively. The easiest way to see that is that

\phi(1_G) = \phi(1_G)\phi(1_G) = \phi(1_G)\phi(1_G)

, so it is getting cancelled. Secondly, let

x,y \in \text{im}(\phi)

. We need to show that

xy^{-1} \in \text{im}(\phi)

. Let

a,b \in G

such that

\phi(a) = x

and

\phi(b) = y

, and we know that

ab^{-1} \in G

. Then

\phi(b)^{-1} = y^{-1}

. So we get the following:

\begin{align*} xy^{-1} &= \phi(a)\phi(b)^{-1} \\ &= \phi(a)\phi(b^{-1}) \\ &= \phi(ab^{-1}). \end{align*}

Therefore, this means that

xy^{-1} \in \text{im}(\phi)

. So

\text{im}(\phi)

is a subgroup of

H

by the subgroup criterion.

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

image of a group homomorphism is a subgroup Archives - Epsilonify

The image of a group homomorphism is a subgroup