Sum of All Ring Products is Associative

Theorem

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

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

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:

$\paren {S T} U = S \paren {T U}$

We have by definition that $S T$ is made up of all finite sums of elements of the form $s \circ t$ where $s \in S, t \in T$.

Therefore, so are $\paren {S T} U$ and $S \paren {T U}$.

Let $z \in \paren {S T} U$.

Then $z$ is a finite sum of elements in the form $x \circ u$ where $x \in ST$ and $u \in U$.

So $x$ is a finite sum of elements in the form $s \circ t$ where $s \in S, t \in T$.

Therefore $z$ is a finite sum of elements in the form $\paren {s \circ t} \circ u$ where $s \in S, t \in T, u \in U$.

As $\struct {R, +, \circ}$ is a ring, $\circ$ is associative.

So $z$ is a finite sum of elements in the form $s \circ \paren {t \circ u}$ where $s \in S, t \in T, u \in U$.

So these elements all belong to $S \paren {T U}$.

Since $S \paren {T U}$ is closed under addition, $z \in S \paren {T U}$.

So:

$\paren {S T} U \subseteq S \paren {T U}$

By a similar argument in the other direction:

$S \paren {T U} \subseteq \paren {S T} U $

and so by definition of set equality:

$\paren {S T} U = S \paren {T U}$

$\blacksquare$