## Theorem

Let $\left({G, +_G}\right)$ be an abelian group whose identity is $e$.

Let $\left({R, +_R, \times_R}\right)$ be a ring whose zero is $0_R$.

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

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

Let $\left \langle {x_m} \right \rangle$ be a sequence of elements of $G$.

Let $\left \langle {\lambda_m} \right \rangle$ be a sequence of elements of $R$ i.e. scalars.

Then:

### Scalar Product with Identity

$\lambda \circ e = 0_R \circ x = e$

### Scalar Product with Inverse

$\lambda \circ \left({- x}\right) = \left({- \lambda}\right) \circ x = - \left({\lambda \circ x}\right)$

### Scalar Product with Sum

$\displaystyle \lambda \circ \left({\sum_{k \mathop = 1}^m x_k}\right) = \sum_{k \mathop = 1}^m \left({\lambda \circ x_k}\right)$

### Product with Sum of Scalar

$\displaystyle \paren {\sum_{k \mathop = 1}^m \lambda_k} \circ x = \sum_{k \mathop = 1}^m \paren {\lambda_k \circ x}$

### Scalar Product with Product

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