Sum of All Ring Products is Additive Subgroup

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

Let $\struct {S, +}$ and $\struct {T, +}$ be additive subgroups of $\struct {R, +, \circ}$.

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

Let $S T$ be defined as:
 * $\ds S T = \set {\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \closedint 1 n}$

Then both $S + T$ and $S T$ are additive subgroups of $\struct {R, +, \circ}$.

Proof
As $\struct {R, +}$ 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 $\struct {R, +, \circ}$.

Let $x, y \in S T$.

We have that $\struct {S T, +}$ is closed.

So $x + y \in S T$.

So, if $\ds y = \sum s_i \circ t_i \in S T$, it follows that:
 * $\ds -y = \sum \paren {-s_i} \circ t_i \in S T$

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