Characterization of subgroup of neutral elements of reducible multiary group

Statement

Consider a reducible multiary group, i.e., a $n$-ary group $G$ with multiplication $f:G^{n}\to G$ such that there exists a group structure on $G$ such that $f(a_{1},a_{2},\dots ,a_{n})=a_{1}a_{2}\dots a_{n}$ for all (possibly repeated) $a_{1},a_{2},{\dot {,}}a_{n}\in G$, with the multiplication on the right being as per the group structure.

Let $C$ be the subset of $G$ comprising those elements that are neutral for $f$. Consider $C$ as a subset of $G$ with its (ordinary 2-ary) group operation. The claim is that $C$ is a subgroup of $G$, and in fact, it is the characteristic subgroup comprising those elements in the center of $G$ whose order divides $n-1$.

Note

Note that due to the equivalence of definitions of reducible multiary group, we know that neutral elements exist if and only if the $n$-ary group is reducible. Thus, for irreducible $n$-ary groups, we would get an empty set instead of a subgroup.

Related facts

• Equivalence of definitions of reducible multiary group
• Groups giving same reducible multiary group are isomorphic

Proof

Given: Group $G$ with identity element $e$. $C$ is the subset of $G$ comprising those elements $u$ of $G$ for which $u^{i}au^{n-1-i}=a$ for all $a\in G$ and all $i\in \{0,1,\dots ,n-1\}$.

To prove: $C$ is precisely the subgroup of $G$ comprising the elements $u$ in the center of $G$ with $u^{n-1}=e$.

Proof:

Step no.	Assertion/construction	Explanation
1	Every element of $C$ is in the center of $G$.	Set $i=0$ to get $au^{n-1}=a$. Now set $i=1$ to get $uau^{n-2}=a$. Equating the two left sides gives $au^{n-1}=uau^{n-2}$. Cancel $u^{n-2}$ from the right of both expressions to get $au=ua$ for all $a\in G$.
2	For all $u\in C$, $u^{n-1}=e$.	Set $i=0$ and $a=e$ to get $eu^{n-1}=e$ which simplifies to $u^{n-1}=e$.
3	If $u$ is in the center of $G$ and $u^{n-1}=e$, then $u\in C$.	Consider the product $u^{i}au^{n-1-i}$. Because $u$ is central, we can commute it past $a$ and get $u^{n-1}a$, which simplifies to $a$ since $u^{n-1}=e$. This is true for all $a\in G$ and all $i\in \{0,1,\dots ,n-1\}$. Thus, $u\in C$.
4	The set of elements $u$ such that $u$ is in the center of $G$ and $u^{n-1}=e$ forms a characteristic subgroup of $G$.	The center is already a subgroup, so we need only check that the condition $u^{n-1}=e$ is a subgroup of it. Since abelian implies universal power map is endomorphism, it can be described as the kernel of the $(n-1)$-power map, which is a subgroup. It is characteristic because of the unique manner of definition.
5	We are done	Steps (1) and (2) show one direction, Step (3) shows the reverse direction. Step (4) confirms that it is a subgroup.