# Groups giving same reducible multiary group are isomorphic

## Statement

Suppose $G$ is a $n$-ary group (i.e., a multiary group with arity $n$) with $n$-ary operation $f$ that is reducible. Suppose $*$ and $\cdot$ are multiplications on $G$, both making $G$ into a group, that both induce the $n$-ary operation $f$ when thought of the usual way:

$f(a_1,a_2,\dots,a_n) = a_1 * a_2 * \dots * a_n = a_1 \cdot a_2 \cdot \dots \cdot a_n$ for all $a_1,a_2,\dots,a_n \in G$

Then, the groups $(G,*)$ and $(G,\cdot)$ are isomorphic groups, and the isomorphism can be expressed explicitly in either group in terms of multiplication by a central element of order dividing $n - 1$.

## Facts used

1. Characterization of subgroup of neutral elements of reducible multiary group

## Proof

Given: Set $G$ with two group operations $*$ and $\cdot$ such that:

$a_1 * a_2 * \dots * a_n = a_1 \cdot a_2 \cdot \dots \cdot a_n$ for all $a_1,a_2,\dots,a_n \in G$

To prove: $(G,*)$ and $(G,\cdot)$ are isomorphic groups

Proof: For simplicity, we will denote $\cdot$ by concatenation but write $*$ explicitly. Let $e$ be the identity element for $\cdot$ and $u$ be the identity element for $*$. $u$ is a neutral element for $f$, hence by Fact (1), it lies in the center of $G$ and $u^{n-1} = e$. Further, by the equality of operations:

$a * b = a * u * \dots * u * b = au^{n-2}b$

Using that $u^{n-1} = e$ and that $u$ is in the center, this gives:

$a * b = aub$

Multiplying both sides by $u$ (this is with respect to $\cdot$, the default multiplication):

$u(a * b) = (ua)(ub)$

We are now in a position to define the isomorphism $(G,*) \to (G,\cdot)$. The isomorphism is:

$x \mapsto u \cdot x$

The above shows that it is a homomorphism. It is clearly bijective, hence it is an isomorphism.