Symmetry groups corresponding to terms in an algebra

Solution 1:

First, let me note that if $\mathsf{SySp}(\mathcal{A})$ contains any nontrivial group then it contains arbitrarily large finite groups. This is just because if $t(x_1,\dots,x_n)$ has symmetry group $H$, then for any $m$-ary operation $s$ the symmetry group of the term $s(t(x_{11},\dots,x_{1n}),\dots,t(x_{m1},\dots,x_{mn}))$ is at least as large as $H^m$ (since you can apply $H$ to each of the sets of $n$ inner variables independently).

Now let $S$ be an infinite set (actually, probably it suffices for $S$ to just be nonempty) and let $\mathcal{A}$ be the free algebra on $S$ with a commutative binary operation. I claim that $\mathsf{SySp}(\mathcal{A})$ consists of entirely of $2$-groups.

To prove this, note that we can think of elements of $\mathcal{A}$ as isomorphism classes of nonempty finite trees in which all non-leaf nodes have two children (we will call these "binary trees") together with a labelling of the leaves by elements of $S$. Similarly, we can identify a term with a binary tree in which each leaf is labelled by one of our variables, and two terms are equivalent iff the corresponding labelled trees are isomorphic. Fix a term $t$ and let $T$ be the corresponding binary tree (without the labelling). Let $H\subseteq S_n$ be the symmetry group of $t$ and let $G$ be the automorphism group of $T$. Note that a permutation $\sigma\in S_n$ is in $H$ iff there exists an automorphism $f:T\to T$ which sends each leaf labelled $i$ to a leaf labelled $\sigma(i)$. Let $G_0\subseteq G$ be the subgroup consisting of automorphisms $f$ which have this property for some $\sigma\in S_n$. We then have a surjective homomorphism $G_0\to H$ sending $f$ to $\sigma$ (this is well-defined due to the requirement that every variable must actually appear in $t$).

So, to show that $H$ is a $2$-group, it suffices to show that $G_0$ is a $2$-group, and for that, it suffices to show that $G$ is a $2$-group. In other words, it suffices show that the automorphism group of any binary tree is a $2$-group. This is easy by induction on the size of the tree. The base case of a single leaf is trivial. For the induction step, suppose we have a binary tree $T$ consisting of a root with two subtrees $T_0$ and $T_1$ below it, and we already know the automorphism groups of $T_0$ and $T_1$ are $2$-groups. There is a homomorphism $\operatorname{Aut}(T)\to S_2$ which sends an automorphism of $T$ to how it permutes $\{T_0,T_1\}$. The kernel of this homomorphism consists of the automorphisms of $T$ that map each of $T_0$ and $T_1$ to themselves, which is just $\operatorname{Aut}(T_0)\times\operatorname{Aut}(T_1)$. Since both $S_2$ and $\operatorname{Aut}(T_0)\times\operatorname{Aut}(T_1)$ are $2$-groups, it follows that $\operatorname{Aut}(T)$ is a $2$-group.

(More generally, instead of a single commutative binary operation, you could consider an arbitrary collection of operations which have given symmetry groups $G_i$ in their inputs. Then a similar argument should show that if $\mathcal{A}$ is a free algebra on infinitely many generators with respect to such operations, the orders of elements of $\mathsf{SySp}(\mathcal{A})$ can only be divisible by primes that divide some $|G_i|$.)