Groups giving same reducible multiary group are isomorphic

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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

Facts used

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


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.