# Scalar Product with Product

## Theorem

Let $\struct {G, +_G}$ be an abelian group.

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

Let $\struct {G, +_G, \circ}_R$ be an $R$-module.

Let $x \in G, \lambda \in R, n \in \Z$.

Then:

$\lambda \circ \paren {n \cdot x} = n \cdot \paren {\lambda \circ x} = \paren {n \cdot \lambda} \circ x$

## Proof

First let $n = 0$.

The assertion follows directly from Scalar Product with Identity.

Next, let $n > 0$.

The assertion follows directly from Scalar Product with Sum and Product with Sum of Scalar, by letting $m = n$ and making all the $\lambda$'s and $x$'s the same.

Finally, let $n < 0$.

The assertion follows from Scalar Product with Product for positive $n$, Scalar Product with Inverse, and from Negative Index Law for Monoids.

$\blacksquare$