Scalar Product with Product

Theorem
Let $\left({G, +_G}\right)$ be an abelian group.

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

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

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

Then:
 * $\lambda \circ \left({n \cdot x}\right) = n \cdot \left({\lambda \circ x}\right) = \left({n \cdot \lambda}\right) \circ x$

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.