PlanetMath: polyadic semigroup

Recall that a semigroup is a non-empty set, together with an associative binary operation on it. Polyadic semigroups are generalizations of semigroups, in that the associative binary operation is replaced by an associative $n$ -ary operation. More precisely, we have

Definition. Let $n$ be a positive integer at least $2$ . A $n$ -semigroup is a non-empty set $S$ , together with an $n$ -ary operation $f$ on $S$ , such that $f$ is associative: $$f(f(a_1,\ldots, a_n),a_{n+1},\ldots, a_{2n-1})=f(a_1,\ldots, f(a_i,\ldots, a_{i+n-1}), \ldots, f_{2n-1})$$ for every $i\in \lbrace 1,\ldots, n\rbrace$ . A polyadic semigroup is an $n$ -semigroup for some $n$ .

An $n$ -semigroup $S$ (with the associated $n$ -ary operation $f$ ) is said to be commutative if $f$ is commutative. An element $e\in S$ is said to be an identity element, or an $f$ -identity, if $$f(a, e, \ldots, e)=f(e,a,\ldots,e)=\cdots = f(e,e,\ldots,a) = a$$ for all $a\in S$ . If $S$ is commutative, then $e$ is an identity in $S$ if $f(a,e,\ldots,e)=a$ .

Every semigroup $S$ has an $n$ -semigroup structure: define $f:S^n\to S$ by \begin{equation} f(a_1,a_n\ldots,a_n)=a_1 \cdot a_2 \cdots \cdot a_n \end{equation}The associativity of $f$ is induced from the associativity of $\cdot$ .

Definition. An $n$ -semigroup $S$ is called an $n$ -group if, in the equation \begin{equation} f(x_1,\ldots, x_n)=a, \end{equation}any $n-1$ of the $n$ variables $x_i$ are replaced by elements of $G$ , then the equation with the remaining one variable has at least one solution in that variable. A polyadic group is just an $n$ -group for some integer $n$ .

$n$ -groups are generalizations of groups. Indeed, a $2$ -group is just a group.

Proof. Let $G$ be a $2$ -group. For $a,b\in G$ , we write $ab$ instead of $f(a,b)$ . Given $a\in G$ , there are $e_1,e_2\in G$ such that $ae_1=a$ and $e_2a=a$ . In addition, there are $x,y\in G$ such that $xa=e_2$ and $ay=e_1$ . So $e_2=xa = x(ae_1)=(xa)e_1= e_2e_1=e_2(ay)=(e_2a)y=ay = e_1$ .

Next, suppose $ae_1=ae_3=a$ . Then the equation $e_2a=a$ from the previous paragraph as well as the subsequent discussion shows that $e_1=e_2=e_3$ . This means that, for every $a\in G$ , there is a unique $e_a\in G$ such that $e_a a =a e_a =a$ . Since $e_a^2 a = e_a (e_a a)=e_a a = a = a e_a = (a e_a) e_a = a e_a^2$ , we see that $e_a$ is idempotent: $e_a^2=e_a$ .

Now, pick any $b\in G$ . Then there is $c\in G$ such that $b=ce_a$ . So $be_a = (c e_a)e_a = c e_a^2= ce_a = b$ . From the last two paragraphs, we see that $e_a = e_b$ . This shows that there is a $e\in G$ such that $ae=ea=a$ for all $a\in G$ . In other words, $e$ is the identity with respect to the binary operation $f$ .

Finally, given $a\in G$ , there are $b,c\in G$ such that $ab=ca=e$ . Then $c = ce = c(ab)= (ca)b=eb = b$ . In addition, if $ab_1=ab_2=e$ , then, from the equation $ca=e$ , we get $b_1=c=b_2$ . This shows $b$ is the unique inverse of $a$ with respect to binary operation $f$ . Hence, $G$ is a group. $\qedsymbol$

Every group has a structure of an $n$ -group, where the $n$ -ary operation $f$ on $G$ is defined by the equation (1) above. Interestingly, Post has proved that, for every $n$ -group $G$ , there is a group $H$ , and an injective function $\phi:G\to H$ with the following properties:

From Post's result, one has the following corollary: an $n$ -semigroup $G$ is an $n$ -group iff equation (2) above has exactly one solution in the remaining variable, when $n-1$ of the $n$ variables are replaced by elements of $G$ .

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