Let a semigroup satisfy $F(x,x)\approx x$, where $F$ is a binary operation symbol.Let $B$ satisfy $x(yz)\approx (xy)z$ and $xx\approx x$. Does $B$ satisfy $F$ as a hyperidentity?We need only consider 6 binary terms $x,y,xy,yx,xyx,yxy$ How do we see that this gives us 6 equations all of the form $$x^a\approx x ?$$ where $a$ is a natural number. What are these equations? The definition of hyperidentity follows:

Let $\sigma:\{f_i:i\in I\}\to W_\tau(X)$ be a mapping assigning to every $n_i$-ary operation symbol $f_i$ of type $\tau$ an $n_i$-ary term, $\sigma(f_i)$. Any such mapping $\sigma$ will be called a hypersubstitution of type $\tau$.

Here $W_\tau(X)$ is the usual recursive definition of terms:

$x_1,...,x_n$ are $n$-ary terms

if $w_1,...,w_m$ are $n$-ary terms and $m=n_i$ (for some $i\in I$) then $f_i(w_1,...,w_m)$ is an $n$-ary term.

NOW we can think of any hypersubstitution $\sigma$ as mapping the term $f_i(x_1,...,x_{n_i})$ to the term $\sigma(f_i)$. It follows that every hypersubstitution of type $\tau$ induces a mapping $\hat{\sigma}:W_\tau(X)\to W_\tau(X)$ as follows:for any $w\in W_\tau(X)$, the term $\hat{\sigma}[w]$ is defined by

(1) $\hat{\sigma}[x]:=x$ for any variable $x\in X$

(2) $\hat{\sigma}[f_i(w_1,...,w_{n_i})]:=\sigma(f_i)(\hat{\sigma}[w_1],...,\hat{\sigma}[w_{n_i}]).$

  • You say that the semigroup satisfies $F(x,x) \approx x$, where $F$ is a binary operation. But what operation? Is it the semigroup operation itself? Then you tell us that it satisfies two more identities, but one of them is trivial, since $B$ is a semigroup (thus, associative). So is $B$ an idempotent semigroup? And the question would be whether or not it hyper-satisfies the idempotent law? And then I suppose that the rest is a hint. – amrsa Feb 8 at 18:57
  • Yes, the F is the semigroup operation itself.And yes, it is idempotent.And finally yes, the question is whether it hyper-satisfies the idempotent law.And the result they claim in the book is that it does satisfy this as a hyper-identity.I just wanted to see how to derive these equations to be satisfied,step by step. – user122424 Feb 9 at 17:17
  • @amrsa PLEASE have a look at this – user122424 Feb 25 at 18:26
up vote 1 down vote accepted

The reason for which it is enough to consider the six binary terms $$t_1(x,y) = x, \; t_2(x,y) = y, \; t_3(x,y) = xy, \; t_4(x,y) = yx, \; t_5(x,y) = xyx, \; t_6(x,y) = yxy$$ is that these form the free band on two generators, $\mathbf F$, and since your interested in an equation on two variables, that's all you need to check.

Now, the result follows from hyper-substituting each of the $t_i$ for $F$ in the equation $F(x,y) \approx x$.
This yields $$t_1(x,x) = t_2(x,x) = x,$$ $$t_3(x,x) = t_4(x,x) = x^2 \approx x,$$ $$t_5(x,x) = t_6(x,x) = x^3 \approx x.$$ The identities $x \approx x^2 \approx x^3$ follow from the fact that $\mathbf B$ satisfies $F(x,x) \approx x$, by hypothesis.

  • How do I know that the hyperidentity satisfied by $t_3(x,y)$ say, is $t_3(x,x)$? – user122424 Feb 10 at 12:24
  • @user122424 That's not the point. The identity you're checking if bands hyper-satisfy is $F(x,x)\approx x$, while the operation is $F(x,y)$; likewise, you don't need that $t(x,y)\approx x$, for every binary term $t$. Actually that wouldn't hold. By hyper-substituting you also check if $t(x,x) \approx x$. Compare with other cases: in hyper-associativity you also only had to check the identities with the relevant variables, not all combinations. – amrsa Feb 10 at 12:35
  • @user122424 Put in a different way (perhaps the previous comment was not very clear), to check whether or not a semigroup is idempotent, you don't check if it satisfies $x \cdot y \approx x$, but just $x \cdot x \approx x$. – amrsa Feb 10 at 12:38
  • That's clear that we just check $x\cdot x \approx x$. But I do not understand why for hyper-satisfaction for $F(x,x)\approx x$ we need say,$t_3(x,x)\approx x$??And not some other left part besides $t_3(x,x)$?I.e. how (2) of my OP applies? – user122424 Feb 10 at 13:15
  • You actually only wrote the definition of hyper-substitution, not of hyper-identity, but that's close. So what's missing is that $u\approx v$ is a hyper-identity if for every hyper-substitution $\sigma$, we have $\hat{\sigma}[u]\approx\hat{\sigma}[v]$. In this case the identity is $F(x,x)\approx x$ and the hyper-substitutions are (1) $\hat{\sigma}_i[x]=x$, $\hat{\sigma}_i[y]=y$ and (2) $\hat{\sigma}_i[F(x,y)]=t_i(x,y)$. In particular, for $t_3$, from $F(x,x)\approx x$ it becomes $t_3(x,x)\approx x$, which is true since $t_3(x,x)=x\cdot x\approx x$. – amrsa Feb 10 at 14:26

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