Let $V$ be a hyperasociative semigroup variety. For hyperasociativiy see below. Then $V$ satisfies the following identity: $$x^2 \approx x^4.$$ A proof attempt is given here: If $V$ is idempotent (i.e. it satisfies $x^2\approx x$) then $x^4 \approx x^2$ is an identity.This is easy. From now on, it is unclear: If not, then $x^2 \not\approx x$, and using $g(x,y)=x^2$ for $F$ in the hyperassociative law forces $x^4\approx x^2$ in $V$.Why? End of the unclear proof.

The definition of hyperassociativity follows.But first is the general hypersubstitution notion.

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

Hyperassociative semigroup variety is then a semigroup variety which satisfies the usual associative law as a hyperidentity.

up vote 1 down vote accepted

I suppose you read that proof in some document (book, article, etc).
Perhaps this is a bit out of context and that justifies what looks like some odd bits in the proof.
I'm referring to the part: "If not, then $x^2 \not\approx x^4$, ..."
Otherwise, it seems OK. Here's why.

The associative identity is $$F(x,F(y,z)) \approx F(F(x,y),z),$$ where $F$ denotes the product in prefix notation.
Hyper-substituting $g$ for $F$ yields $$g(x,g(y,z)) \approx g(g(x,y),z),$$ whence $$x^2 \approx g(x^2,z),$$ and finally, $$x^2 \approx (x^2)^2 \approx x^4.$$

  • They do not write "If not, then $x^2 \not \approx x^4$ but "If not, then $x^2\not\approx x$. They somewhere need that $x^2$ is not equal ($\not\approx$) to the variable $x$. – user122424 Feb 1 at 21:36
  • Ah, sorry I misread it! Then I suppose it is suppose to mean that, if the semigroup is idempotent then it's easy to see that $x^2 \approx x^4$ (actually, $x^n \approx x^m$, for all $n,m \geq 1$); if it isn't, but it hyper-satisfies the associative law, then it satisfies $x^2 \approx x^4$, just the same. Does this make sense? – amrsa Feb 1 at 21:37
  • Yes. But I wonder whether they actually used what they write:"If not, then $x^2\not\approx x$, and using $g(x,y)=x^2$ for $F$....? Is this explicitly used?I think there may be some issue with a hypersubstitution for a variable. – user122424 Feb 1 at 21:41
  • Well, If $V$ is idempotent, it satisfies $x^2 \approx x$ and then $x^2 \approx x^4$; if it is not idempotent, using the term $g(x,y) = x^2$ for $F$... Isn't that what I did in my answer? I mean, I suppose that what I wrote is what they meant, although they might have found that they could skip the details for whatever reason... – amrsa Feb 1 at 21:47
  • So may be there is no other hidden meaning in that "If not, then $x^2\not\approx x$..."? OK then. – user122424 Feb 1 at 21:56

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