Tour:Equality of left and right neutral element

This article adapts material from the main article: equality of left and right neutral element

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WHAT YOU NEED TO DO:

• Read, and understand, the statement below, and try proving it.
• Understand clearly why this statement implies the first stated corollary: that the neutral element of a magma is determined by its binary operation.
• Go through the proof, and make sure you understand it.

Statement

Let $S$ be a magma (set with binary operation). Suppose $e_1$ is a left neutral element of $S$, and $e_2$ is a right neutral element.

Here, $e_1$ is a left neutral element if $e_1 * a = a$ for all $a \in S$, and $e_2$ is a right neutral element if $a * e_2 = a$ for all $a \in S$.

Then, $e_1 = e_2$, and it is therefore a (two-sided) neutral element.

Corollaries

• Binary operation on magma determines neutral element: If there exists a neutral element (i.e., an element that is simultaneously left and right neutral), then it is unique, and is determined by the binary operation.
• If there exists a left neutral element, then there can exist at most one right neutral element, and they must be equal.
• If there exist two different left neutral elements, there cannot exist any right neutral element.

Proof

Proof idea

A left neutral element is an element that is recessive when placed on the left (in other words, it gives way to the element on its right). A right neutral element is recessive when placed on the right (it gives way to the element on its left). By pitting these elements against each other, we force both of them to give way to each other, which forces them to be equal.

Formal proof

Given: A magma $S$ with binary operation $*$. $e_1 \in S$ is a left neutral element for $*$, i.e., $e_1 * a = a \ \forall \ a \in S$. $e_2 \in S$ is a right neutral element for $*$, i.e., $a * e_2 = a \ \forall \ a \in S$.

To prove: $e_1 = e_2$

Proof: Consider the product $e_1 * e_2$. This is equal to $e_2$ (because $e_1$ is left neutral) and is also equal to $e_1$ (because $e_2$ is right neutral). Hence, $e_1 = e_2$.

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