Image of Submodule under Linear Transformation is Submodule

Theorem
Let $\struct {R, +_R, \times_R}$ be a ring.

Let $\struct {G, +_G, \circ_G}_R$ and $\struct {H, +_H, \circ_H}_R$ be $R$-modules.

Let $\phi: G \to H$ be a linear transformation.

Let $M$ be a submodule of $G$.

Then $\phi \sqbrk M$ is a submodule of $H$.

Proof
Let $N = \phi \sqbrk M$ be the image set of $M$ under $\phi$.

By definition, a linear transformation $\phi: G \to H$ is, in particular, a (group) homomorphism from the group $\struct {G, +_G}$ to the group $\struct {H, +_H}$.

We have that $M$ is a submodule of $G$.

So from Elements of Submodule form Subgroup, $M$ forms a subgroup of $G$.

From Group Homomorphism Preserves Subgroups, $N$ is therefore a subgroup of $H$.

It remains to be shown that $N$ is closed for scalar product:
 * $\forall \lambda \in R, x \in N: \lambda \circ_H x \in N$

As $M$ is a submodule of $G$, $M$ itself is closed for scalar product:
 * $\forall \lambda \in R, y \in N: \lambda \circ_G y \in M$

We have: