Equivalence of definitions of reducible multiary group

This article gives a proof/explanation of the equivalence of multiple definitions for the term reducible multiary group
View a complete list of pages giving proofs of equivalence of definitions

Statement

Recall that a multiary group is a $n$-ary group for some $n \ge 2$. The following are equivalent for a $n$-ary group with multiplication $f:G^n \to G$:

1. There exists a group structure on $G$ such that $f(a_1,a_2,\dots,a_n) = a_1a_2\dots a_n$ 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. For more on this, see group is n-ary group for all n
2. There exists a neutral element $e \in G$ for $f$: $f$ evaluated at any tuple where $n - 1$ of the entries are equal to $e$ and the remaining entry is $a \in G$, gives output $a$. This is true regardless of where we place $a$ and is also true if $a = e$.

Related facts

• Characterization of subgroup of neutral elements of reducible multiary group
• Groups giving same reducible multiary group are isomorphic

Proof

(1) implies (2)

This is immediate: we can take $e$ to be the neutral element (i.e., the identity element) of $G$ as a group.

(2) implies (1)

Given: A set $G$, function $f:G^n \to G$ making it a $n$-ary group, an element $e$ that is neutral for $f$.

To prove: There exists a group structure on $G$ with respect to which $f(a_1,a_2,\dots,a_n) = a_1a_2 \dots a_n$ for all $a_1,a_2,\dots,a_n \in G$.

Proof: We use the notation $e_i$ to denote $e$ written in succession $i$ times.

Step no.	Assertion/construction	Given data used	Previous steps used	Explanation
1	$f(x,y,e_{n-2}) = f(e_i,x,e_j,y,e_k)$ for all $x,y \in G$ and all $i,j,k \ge 0$ with $i + j + k = n - 2$.	$e$ is neutral for $f$, $f$ satisfies associativity becaus it gives a $n$-ary group operation		By asociativity, we have $f(e_{k+1},f(e_i,x,e_j,y,e_k),e_{i+j}) = f(f(e_{k+1+i},x,e_j),y,e_{n-2})$. Using that $e$ is neutral, the outer $f$ of the left side vanishes and the left side becomes $f(e_i,x,e_j,y,e_k)$. On the right side, using that $f$ is neutral, the inside $f$ vanishes and we are left with $f(x,y,e_{n-2})$, completing the proof.
2	We can define an operation $xy := f(x,y,e_{n-2})$, and this equals $f$ evaluated on any tuple with $(n-2)$ $e$s, one $x$ and one $y$, with the $x$ appearing to the left of the $y$.		Step (1)	Step-direct
3	The operation defined in Step (2) is associative, i.e., $(ab)c = a(bc)$ for all $a,b,c \in G$	$f$ satisfies $n$-ary associativity	Step (2)	We write the operation as $xy := f(x,e_{n-2},y)$. We thus get$(ab)c = f(f(a,e_{n-2},b),e_{n-2},c) = f(a,e_{n-2},f(b,e_{n-2},c)) = a(bc)$
4	The operation defined in Step (2) has identity element $e$	$e$ is neutral for $f$	Step (2)	$e$ works as an identity element: for all $a \in G$, $ae = f(a,e_{n-1}) = a$. Similarly, $ea = f(e_{n-1},a) = a$.
5	The operation defined in Step (2) admits two-sided inverses		Steps (2), (3), (4)	We show one-sided inverses from the unique solution condition: the right inverse of $a$ is the unique solution $b$ to $f(a,e,\dots,e,b) = e$. Similarly, the left inverse of $a$ is the unique $c$ such that $f(c,e,\dots,e,a) = a$. Now, we use the equality of left and right inverses in monoid to conclude that two-sided inverses exist.
6	The operation defined in Step (2) gives a group structure on $G$		Steps (2), (3), (4), (5)	Step-combination direct
7	For any elements $a_1,a_2,\dots,a_i \in G$ with $i \le n$, $f(a_1,a_2,\dots,a_i,e_{n-i}) = a_1a_2\dots a_i$.	$f$ is associative	Steps (2), (6)	We prove this by induction on $i$. Assuming true for $i - 1$, write $f(f(a_1,a_2,\dots,a_i,e_{n-i}),e_{n-1}) = f(a_1,f(a_2,\dots,a_i,e_{n-i+1})e_{n-2})$. The left side simplifies to the desired left side. On the right side, the second input simplifies to $a_2\dots a_i$ by inductive hypothesis, so now using the definition of group product, the product simplifies to $a_1a_2 \dots a_i$ as desired.
8	We are done		Steps (2), (6), (7)	By Steps (2) and (6), we have a group structure on $G$, which setting $i = n$ in Step (7), gives the desired $n$-ary group, completing the proof.

Caution about non-uniqueness of neutral elements

For a reducible multiary group, there may be many different choices of neutral element. Each choice yields a potentially different binary operation, though the group structures we obtain are isomorphic. For instance, for the $n$-ary group obtained from the cyclic group of order $n - 1$, every element is a neutral element. The different group structures correspond to affine shifts in the original group, i.e., relabeling all elements by adding a group element to them.