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:

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:

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:

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:

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