# Middle-associative elements of magma form submagma

## Statement

Let $(S,*)$ be a magma (a set $S$ with binary operation $*$). Call an element $b \in S$ a middle-associative element (or middle nuclear element) if the following holds: $a * (b * c) = (a * b) * c \ \forall \ a,c \in S$

Then:

1. The set of middle-associative elements of $S$ forms a submagma of $S$ that is in fact a semigroup under the induced operation. This submagma is termed the middle nucleus of $S$.
2. If $S$ contains a neutral element $e$, then $e$ is middle-associative, and is a neutral element for the submagma of middle-associative elements.
3. If $S$ contains a nil element $n$, then $n$ is middle-associative, and is a nil element for the submagma of middle-associative elements.

## Related facts

All these proofs make crucial use of the associativity pentagon: the pentagon describing the relation between the five different ways of associating a product of length four.

## Proof

### Proof idea

The idea behind the proof is the associativity pentagon, which states that the five different ways of parenthesizing an expression of length four form a cyclic pentagon, with each expression related to exactly two others via a single application of the associativity law. In order to move along one edge of the pentagon, we instead move along the path comprising the remaining four edges, and use the fact that each step there is allowed because one of $b_1$ and $b_2$ is the middle term in each of the applications of the law.

### Proof of (1)

Given: A magma $(S,*)$, two middle-associative elements (i.e., middle nuclear) $b_1,b_2 \in S$

To prove: $b_1 * b_2$ is middle-associative (i.e., middle nuclear)

Proof: We need to show that, for any $a, c \in S$, we have: $a * ((b_1 * b_2) * c) = (a * (b_1 * b_2)) * c$

Let's do this. Start with the left side and proceed as follows: $a * ((b_1 * b_2) * c) = a * (b_1 * (b_2 * c)) = (a * b_1) * (b_2 * c) = ((a * b_1) * b_2) * c = (a * (b_1 * b_2)) * c$.

In two of the steps, we use that $b_1$ is middle-associative (i.e., middle nuclear), and in two of the steps, we use that $b_2$ is middle-associative (i.e., middle nuclear).

Here is a diagrammatic depiction of the proof: