Say that a clone (in the sense of universal algebra) $\mathfrak{C}$ has *almost abelian symmetry groups* (= **aasg**) iff for each function $f(x\_1,...,x\_n)\in\mathfrak{C}$ there is an abelian subgroup $A\subseteq S\_n$ and a set $X\subseteq \{1,...,n\}$ such that $$\langle A\cup \mathsf{Fix}(X)\rangle=\{\sigma: f=\lambda x\_1,...,x\_n. f(x\_{\sigma(1)},...,x\_{\sigma(n)})\}$$ (where "$\mathsf{Fix}(X)$" denotes the subgroup of $S\_n$ consisting of permutations fixing $X$ pointwise and "$\langle U\rangle$" denotes the subgroup of $S\_n$ generated by $U$). Basically, $\mathfrak{C}$ has aasg if every function in $\mathfrak{C}$ has an abelian symmetry group *once* we ignore "dummy variable" issues.


(Note that we're looking at the symmetry groups of the elements of the clone, not the symmetry group of the clone - in any sense - itself.)


Coming from [this old question of mine](https://mathoverflow.net/questions/412563/what-classes-of-groups-can-arise-as-symmetry-groups-of-terms), I'm interested in understanding aasg-ness. It seems natural to start with the case where we look at clones on a $2$-element set, since things are already nontrivial but well-understood there:



> 
> **Question 1**: Which clones on a $2$-element set have aasg?
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Note that [Post's lattice](https://en.wikipedia.org/wiki/Post%27s_lattice#/media/File:Post-lattice.svg) consists of finitely many elements together with eight infinite families, so - since aasg-ness is inherited by subclones - there are only a few different "shapes" that the set of aasg clones could have. In particular, the set of aasg clones on a two-element set is (in the appropriate sense) computable (thanks to Emil Jerabek for pointing this out!).


A follow-up question is whether, like general clones, things "blow up" once we get to three elements:



> 
> **Question 2**: How many aasg clones are there on $\{1,2,3\}$? If there are only countably many, what is the least finite $n$ (if any) such that there are continuum-many aasg clones on $\{1,...,n\}$?
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> 
> 



When does a clone on a two-element set have almost abelian symmetry groups?
**Question 1**:
Is the set of clones on $\{1,2\}$ which have aasg computable?
The answer is Yes. The aasg clones are just the essentially unary clones
on $\{1,2\}$, and there are six of them.
**Easy direction:**
Assume that $\mathfrak C$ is essentially unary
and $f(x\_1,\ldots,x\_n)\in \mathfrak C$
does not depend on its first $n-1$
variables. The symmetry group of $f$ is
$\textrm{Sym}(\{1,\ldots,n-1\})$ if
$f$ depends on its last variable and is
$\textrm{Sym}(\{1,\ldots,n\})$ otherwise.
Both cases satisfy the aasg property.



**Less easy direction:**
Assume that $\mathfrak C$ is a clone on $\{0,1\}$
that is not essentially unary.
[I changed my underlying set to $\{0,1\}$ so that I can
write operations as Boolean expressions.]
I will argue that (i) $\mathfrak C$
contains a totally symmetric operation
of arity $\geq 3$ that depends on all variables, and that (ii)
a clone with a totally symmetric operation
of arity $\geq 3$ that depends on all variables
fails the aasg property.



For (i), just examine Post's lattice.
Every clone that is not essentially unary
contains one of the Boolean operations
(a) $x\wedge y\wedge z$ (ternary conjunction),  

(b) $x\vee y\vee z$ (ternary disjunction),  

(c) $x \oplus y \oplus z$ (ternary addition modulo $2$),  

(d) $(x\wedge y)\vee (x\wedge z)\vee (y\wedge z)$ (ternary majority).
For (ii), assume that $\mathfrak C$ contains
a totally symmetric operation $f(x\_1,\ldots,x\_m)$
which depends on all variables and that $m\geq 3$.
For $n\geq 3$, the symmetry group of
$g(x\_1,\ldots,x\_m,x\_{m+1},\ldots,x\_{m+n}):=f(x\_1,\ldots,x\_m)$
is $S\_m\times S\_n$, and both factors
are nonabelian.
**Claim.** The symmetry group of $g$,
which is $S\_m\times S\_n\leq S\_{m+n}$, cannot have the form
$\langle A\cup \textrm{Fix}(X)\rangle$ for any
abelian subgroup $A\subseteq S\_{m+n}$
and any $X\subseteq \{1,\ldots,m+n\}$.
Assume otherwise.
Note that $\textrm{Fix}(X)$
is generated by the transpositions it contains,
and that this set of transpositions is `connected'
in the sense that $(i\;j), (k\;l)\in\textrm{Fix}(X)$
implies $(j\;k)\in \textrm{Fix}(X)$.
The only connected subsets of transpositions
in $S\_m\times S\_n$ lie in $S\_m\times \{1\}$
or $\{1\}\times S\_n$, so $\textrm{Fix}(X)$
lies in either
$S\_m\times \{1\}$
or $\{1\}\times S\_n$. This is enough to see that
if $N$ is the normal subgroup $S\_m\times S\_n$
that is generated by $\textrm{Fix}(X)$, then
$(S\_m\times S\_n)/N$ is nonabelian,
which can't happen if
$S\_m\times S\_n=\langle A\cup \textrm{Fix}(X)\rangle$.


















