5

Let $\mathfrak{E}=(\mathbb{N};\mathsf{exp})$ be the structure consisting of the natural numbers together with exponentiation. Given a term $t$ in the language of $\mathfrak{E}$ - that is, an expression built only out of variables and the exponentiation function - we can associate to $t$ the group $E_t$ consisting of all permutations of the variables occurring in $t$ which result in the same function when interpreted in $\mathfrak{E}$. So for example, abusing notation in the obvious way, if $t=((x^y)^z)^w$ then $E_t\cong S_3$ since the three variables $y,z,w$ can be freely permuted.

I've asked earlier exactly which groups appear as symmetry groups of terms in $\mathfrak{E}$. However, that seems harder than I originally expected. I think the following special case is more likely to have a quick answer:

Is there a term $t$, using exponentiation alone, with $E_t\cong C_3$ (or indeed any cyclic group $C_n$ for $n>2$)?

I suspect the answer is negative, but I don't immediately see how to prove this. Note that the equational theory of this structure is generated by the single nontrivial equation $(x^y)^z=(x^z)^y$; see here.

Noah Schweber
  • 245,398
  • 1
    I am going use the notation $y \to x$ for $x^y$ and $x \to y \to z$ for $x \to (y \to z)$. With these notational conventions in place, have you considered the term $((x \to y) \to z) \to ((y \to z) \to x) \to ((z \to x) \to y) \to w$? It seems to me that this term might have the required symmetry group. – Pilcrow Jan 17 '22 at 00:39
  • @Pilcrow Bringing multiplication in for clarity, your term is equivalently written as $$w^{[z^{(y^x)}]\cdot[x^{(z^y)}]\cdot [y^{(x^z)}]},$$ right? – Noah Schweber Jan 17 '22 at 01:44
  • Yes, that's right. – Pilcrow Jan 17 '22 at 02:06
  • @Pilcrow That seems to work, and also to get all $C_n$s, e.g. $C_4$ arises from $$w^{[a^{b^{c^d}}]\cdot[b^{c^{d^a}}]\cdot[c^{d^{a^b}}]\cdot [d^{a^{b^c}}]},$$ etc. I feel a bit silly in retrospect. If you add this as an answer I'll accept it! – Noah Schweber Jan 17 '22 at 03:17
  • Actually this seems much more broadly useful. Let $[x_1,...,x_k]$ be shorthand for $x_1^{(x_2^{(...^{x_k})})}$. For $G$ a subgroup of $S_k$, let $t_G$ be the term $$w^{\prod_{\sigma\in G}[x_{\sigma(1)},..., x_{\sigma(k)}]}$$ (with $w, x_1,...,x_k$ distinct variables). Then I think we get $E_{t_G}\cong G$, right? – Noah Schweber Jan 17 '22 at 03:22
  • 1
    Yes, I agree that this should work. The reason that I made a comment rather than an answer is that I am not sure how to easily prove that there are no symmetries other than the obvious ones. Someone more at home in number theory would probably know which theorem to pull off the shelf for this purpose. If you know how to write down a proper proof, please feel to transform my comment into a proper answer. – Pilcrow Jan 17 '22 at 14:24

0 Answers0