Equality of left and right neutral element

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.

Similar facts
There are many statements of a similar form to this: every left-based construction is equal to every right-based construction. A similar flavor is observed in max-min theorems.

Category:Statements of left-right equivalence

Here is a tabulation of some closely related statements:

Opposite facts

 * It is possible to construct a magma with more than one left neutral element, but such a magma must have no right neutral element. An extreme example of this is a magma with the multiplication defined as $$a * b := b$$ for all $$a,b$$ in the semigroup. For such a magma, every element is left neutral. if the magma has size two or more, there is no right neutral element. Note that any such magma is a semigroup.
 * There is a corresponding definition of neutral element for a multiary operation. For an operation with $$n$$ inputs, we can define a neutral element with respect to a given position $$i$$ as an element $$e$$ such that, when $$e$$ is placed in all positions other than $$i$$, the output is the $$i^{th}$$ input. A neutral element is an element that is neutral with respect to all positions. There is no easy analogous statement for multiary operations; in fact, even very nicely behaved multiary operations can have multiple neutral elements.

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, $$e_1 = e_2$$.

Proof in multiplication table terms
We use the same notation as in the formal proof.

Consider the multiplication table of $$S$$. Imagine that the multiplication table is written with the row element multiplied on the left and the column element multiplied on the right. The condition that $$e_1$$ is left neutral means that the row corresponding to $$e_1$$ is a replica of the column headers row. The condition that $$e_2$$ is right neutral means that the column corresponding to $$e_2$$ is a replica of the row headers column. Now, we look at the cell of the multiplication table corresponding to the product $$e_1 * e_2$$. This must equal both the column header $$e_2$$ and the row header $$e_1$$, hence $$e_1 = e_2$$.