# Homomorphic Image of R-Module is R-Module

## Theorem

Let $\left({R, +_R, \times_R}\right)$ be a ring.

Let $\left({G, +_G, \circ_G}\right)_R$ be an $R$-module.

Let $\left({H, +_H, \circ_H}\right)_R$ be an $R$-algebraic structure.

Let $\phi: G \to H$ be a homomorphism.

Then the homomorphic image of $\phi$ is an $R$-module.

## Proof

Let us write $\phi \left({G}\right)$ to denote the homomorphic image of $\phi$.

From Image of Group Homomorphism is Subgroup, $\phi \left({G}\right)$ is a subgroup of $\left({H, +_H}\right)$.

For any $\phi \left({g}\right)$ and $\phi \left({g'}\right)$ in $\phi \left({G}\right)$, we have:

 $\displaystyle \phi \left({g}\right) +_H \phi \left({g'}\right)$ $=$ $\displaystyle \phi \left({g +_G g'}\right)$ $\phi$ is a homomorphism $\displaystyle$ $=$ $\displaystyle \phi \left({g' +_G g'}\right)$ $\left({G, +_G}\right)$ is an abelian group $\displaystyle$ $=$ $\displaystyle \phi \left({g'}\right) +_H \phi \left({g}\right)$ $\phi$ is a homomorphism

hence $\phi \left({G}\right)$ is an abelian group.

Now we can turn to showing that $\phi \left({G}\right)$ is an $R$-module.

To do this, we take the $R$-module axioms in turn.

### Proof of $(1)$

It is to be shown that for all $\lambda \in R$ and $\phi \left({g}\right), \phi \left({g'}\right) \in \phi \left({G}\right)$:

$\lambda \circ_H \left({\phi \left({g}\right) +_H \phi \left({g'}\right)}\right) = \left({\lambda \circ_H \phi \left({g}\right)}\right) +_H \left({\lambda \circ_H \phi \left({g'}\right)}\right)$

Compute, using that $\phi$ is a homomorphism repetitively:

 $\displaystyle \lambda \circ_H \left({\phi \left({g}\right) +_H \phi \left({g'}\right)}\right)$ $=$ $\displaystyle \lambda \circ_H \phi \left({g +_G g'}\right)$ $\displaystyle$ $=$ $\displaystyle \phi \left({\lambda \circ_G \left({g +_G}\right)}\right)$ $\displaystyle$ $=$ $\displaystyle \phi \left({\left({\lambda \circ_G g}\right) +_G \left({\lambda \circ_G g'}\right)}\right)$ $G$ is an $R$-module $\displaystyle$ $=$ $\displaystyle \phi \left({\lambda \circ_G g}\right) +_H \phi \left({\lambda \circ_G g'}\right)$ $\displaystyle$ $=$ $\displaystyle \left({\lambda \circ_H \phi \left({g}\right)}\right) +_H \left({\lambda \circ_H \phi \left({g'}\right)}\right)$

$\Box$

### Proof of $(2)$

It is to be shown that for all $\lambda, \mu \in R$ and $\phi \left({g}\right) \in \phi \left({G}\right)$:

$\left({\lambda +_R \mu}\right) \circ_H \phi \left({g}\right) = \left({\lambda \circ_H \phi \left({g}\right)}\right) +_H \left({\mu \circ_H \phi \left({g}\right)}\right)$

Compute, using that $\phi$ is a homomorphism repetitively:

 $\displaystyle \left({\lambda +_R \mu}\right) \circ_H \phi \left({g}\right)$ $=$ $\displaystyle \phi \left({\left({\lambda +_R \mu}\right) \circ_G g}\right)$ $\displaystyle$ $=$ $\displaystyle \phi \left({\left({\lambda \circ_G g}\right) +_G \left({\mu \circ_G g}\right)}\right)$ $G$ is an $R$-module $\displaystyle$ $=$ $\displaystyle \phi \left({\lambda \circ_G g}\right) +_H \phi \left({\mu \circ_G g}\right)$ $\displaystyle$ $=$ $\displaystyle \left({\lambda \circ_H \phi \left({g}\right)}\right) +_H \left({\mu \circ_H \phi \left({g}\right)}\right)$

$\Box$

### Proof of $(3)$

It is to be shown that for all $\lambda, \mu \in R$ and $\phi \left({g}\right) \in \phi \left({G}\right)$:

$\left({\lambda \times_R \mu}\right) \circ_H \phi \left({g}\right) = \lambda \circ_H \left({\mu \circ_H \phi \left({g}\right)}\right)$

Compute, using that $\phi$ is a homomorphism repetitively:

 $\displaystyle \left({\lambda \times_R \mu}\right) \circ_H \phi \left({g}\right)$ $=$ $\displaystyle \phi \left({\left({\lambda \times_R \mu}\right) \circ_G g}\right)$ $\displaystyle$ $=$ $\displaystyle \phi \left({\lambda \circ_G \left({\mu \circ_G g}\right)}\right)$ $G$ is an $R$-module $\displaystyle$ $=$ $\displaystyle \lambda \circ_H \phi \left({\mu \circ_G g}\right)$ $\displaystyle$ $=$ $\displaystyle \lambda \circ_H \left({\mu \circ_H \phi \left({g}\right)}\right)$

$\Box$

Having verified that $\phi \left({G}\right)$ satisfies the three axioms, we conclude it is an $R$-module.

$\blacksquare$