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 $n-1$.

Related facts

• Equivalence of definitions of reducible multiary group
• Characterization of subgroup of neutral elements of reducible multiary group

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.