# Sum of All Ring Products is Additive Subgroup

## Theorem

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

Let $\left({S, +}\right)$ and $\left({T, +}\right)$ be additive subgroups of $\left({R, +, \circ}\right)$.

Let $S + T$ be defined as subset product.

Let $S T$ be defined as:

- $\displaystyle S T = \left\{{\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \left[{1 \,.\,.\ n}\right]}\right\}$

Then both $S + T$ and $S T$ are additive subgroups of $\left({R, +, \circ}\right)$.

## Proof

As $\left({R, +}\right)$ is abelian (from the definition of a ring), we have:

- $S + T = T + S$

from Subset Product of Commutative is Commutative.

So from Subset Product of Subgroups it follows that $S + T$ is an additive subgroup of $\left({R, +, \circ}\right)$.

Let $x, y \in S T$.

We have that $\left({S T, +}\right)$ is closed.

So $x + y \in S T$.

So, if $\displaystyle y = \sum s_i \circ t_i \in S T$, it follows that:

- $\displaystyle -y = \sum \left({-s_i}\right) \circ t_i \in S T$

By the Two-Step Subgroup Test, we have that $S T$ is an additive subgroup of $\left({R, +, \circ}\right)$.

$\blacksquare$

## Sources

- 1970: B. Hartley and T.O. Hawkes:
*Rings, Modules and Linear Algebra*... (previous) ... (next): $\S 2.1$: Subrings: Lemma $2.3 \ \text{(ii)}$