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:

$$f(f(a_1,\mathrm{\dots },a_n),a_{n+1},\mathrm{\dots },a_{2n-1})=f(a_1,\mathrm{\dots },f(a_i,\mathrm{\dots },a_{i+n-1}),\mathrm{\dots },f_{2n-1})$$

for every $i\in \{1,\mathrm{\dots },n\}$. 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

$$f(a,e,\mathrm{\dots },e)=f(e,a,\mathrm{\dots },e)=\mathrm{\cdots }=f(e,e,\mathrm{\dots },a)=a$$

for all $a\in S$. If $S$ is commutative, then $e$ is an identity in $S$ if $f(a,e,\mathrm{\dots },e)=a$.

Every semigroup $S$ has an $n$-semigroup structure: define $f:S^n\to S$ by

$$f(a_1,a_n\mathrm{\dots },a_n)=a_1\cdot a_2\mathrm{\cdots }\cdot a_n$$(1)

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

$$f(x_1,\mathrm{\dots },x_n)=a,$$(2)

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 $a{e}_1=a$ and $e_{2}a=a$. In addition, there are $x,y\in G$ such that $xa=e_2$ and $ay=e_1$. So $e_2=xa=x(a{e}_1)=(xa)e_1=e_2{e}_1=e_2(ay)=(e_{2}a)y=ay=e_1$.

Next, suppose $a{e}_1=a{e}_3=a$. Then the equation $e_{2}a=a$ 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 $b=c{e}_a$. So $b{e}_a=(c{e}_a)e_a=c{e}_a^2=c{e}_a=b$. 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 $a{b}_1=a{b}_2=e$, 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.∎

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 $\varphi :G\to H$ with the following properties:

1. $\varphi (G)$ generates $H$

2. $\varphi (f(a_1,\mathrm{\dots },a_n))=\varphi (a_1)\mathrm{\cdots }\varphi (a_n)$

If we call the group $H$ with the two above properties a covering group of $G$, then Post’s theorem states that every $n$-group has a covering group.

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 $G$.