Neutral element for a multiary operation

Definition

All-sided neutral element

Suppose $S$ is a set and $f:S^{n}\to S$ is a $n$-ary operation for $f$. An element $e\in S$ is termed a neutral element for $f$ if the following holds: $f$ evaluated at any $n$-tuple where $n-1$ of the entries are equal to $e$ and the remaining entry is $a\in S$, gives output $a$. This is true regardless of where we place $a$ and is also true if $a=e$.

The term neutral element, when used without qualification, is used in the context $n=2$, i.e., for a binary operation, i.e., a magma.

Left neutral element

Suppose $S$ is a set and $f:S^{n}\to S$ is a $n$-ary operation for $f$. An element $e\in S$ is termed a left neutral element for $f$ if the following holds: $f(e,e,e,\dots ,e,a)=a$ for all $a\in S$.

Right neutral element

Suppose $S$ is a set and $f:S^{n}\to S$ is a $n$-ary operation for $f$. An element $e\in S$ is termed a right neutral element for $f$ if the following holds: $f(a,e,e,e,\dots ,e)=a$ for all $a\in S$.

Neutral element for a given position

Suppose $S$ is a set and $f:S^{n}\to S$ is a $n$-ary operation for $f$. Suppose $i\in \{1,2,\dots ,n\}$. An element $e\in S$ is termed a neutral element for position $i$ if $f$ if the following holds: $f(e,\dots ,e,a,e,e,\dots ,e)=a$ for all $a\in S$ where $a$ appears in the $i^{th}$ position.

Facts

The case $n=2$ is special because we can deduce equality of left and right neutral element and therefore also deduce that binary operation on magma determines neutral element. For higher $n$, there could be more than one neutral element.