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Let $\mathfrak{A}$ be an algebra (in the sense of universal algebra). To each term $t(x_1,...,x_n)$ in the language of $\mathfrak{A}$ in which each variable actually appears we can assign a group $G_\mathfrak{A}(t)\subseteq S_n$ consisting of all permutations of the variables which results in the same function: $$G_\mathfrak{A}(t)=\{\sigma\in S_n: \forall a_1,...,a_n\in\mathfrak{A}(t(a_1,...,a_n)=t(a_{\sigma(1)},...,a_{\sigma(n)})\}.$$

Now let $\mathbb{G}(\mathfrak{A})$ be the class of isomorphism types of groups of the form $G_\mathfrak{A}(t)$ for some term $t$. I'm curious which classes of groups can arise this way.

Originally I asked for a complete answer to this question, but in retrospect that was overly ambitious. To keep things reasonably answerable, let me restrict to the following natural candidates (playing fast and loose with isomorphism-type-vs.-group issues for simplicity) - I would love an answer, or even partial progress, to either question (or anything similar really, I'm profoundly stuck-in-the-weeds here):

Q1: Is there an $\mathfrak{A}$ such that $\mathbb{G}(\mathfrak{A})$ consists exactly of the finite $p$-groups for some prime $p$?

Certainly it's possible to get only $2$-groups - see e.g. this MSE answer of Eric Wofsey. However, getting exactly the $2$-groups, or $p$-groups for any fixed $p$, seems much harder.

Q2: Is there a $\mathfrak{B}$ such that $\mathbb{G}(\mathfrak{B})$ consists exactly of the finite abelian groups?

I really have no relevant information for this question, but it seems like a natural one to ask.

Note that Keith Kearnes' answer below addresses a variant of this question, in which variable appearance is replaced by variable dependence, and does not seem to immediately generalize to address this version.

Here are some example $\mathbb{G}(\mathfrak{A})$s (ignoring up-to-isomorphism details):

• If $\mathfrak{A}=(A;\star)$ where $\star:A^2\rightarrow A$ is a bijection, then $\mathbb{G}(\mathfrak{A})$ consists only of the trivial group.

• If $\mathfrak{A}=(\mathbb{N};\max)$, then $\mathbb{G}(\mathfrak{A})=\{S_n:n\in\mathbb{N}\}$.

• Although in the natural numbers with exponentiation alone there is essentially only one nontrivial equation, it turns out that every finite group shows up - see here.

• One of the few negative results I know is that the class of finite cyclic groups is not of the form $\mathbb{G}(\mathfrak{A})$ for any $\mathfrak{A}$. To see this, suppose $t(x_1,...,x_k)$ is a $\mathfrak{A}$-term with $G_\mathfrak{A}(t)\cong C_2$. Then the term $$s(x_{1,1},...,x_{k,k}):=t(t(x_{1,1},...,x_{1,k}), ..., t(x_{k,1},...,x_{k,k}))$$ has $G_\mathfrak{A}(s)$ having too many elements of order $2$ to be cyclic. However, this sort of idea doesn't seem to be useful for either Q1 or Q2 above.

Let $\mathfrak{A}$ be an algebra (in the sense of universal algebra). To each term $t(x_1,...,x_n)$ in the language of $\mathfrak{A}$ in which each variable actually appears we can assign a group $G_\mathfrak{A}(t)\subseteq S_n$ consisting of all permutations of the variables which results in the same function: $$G_\mathfrak{A}(t)=\{\sigma\in S_n: \forall a_1,...,a_n\in\mathfrak{A}(t(a_1,...,a_n)=t(a_{\sigma(1)},...,a_{\sigma(n)})\}.$$

Now let $\mathbb{G}(\mathfrak{A})$ be the class of isomorphism types of groups of the form $G_\mathfrak{A}(t)$ for some term $t$. I'm curious which classes of groups can arise this way.

Originally I asked for a complete answer to this question, but in retrospect that was overly ambitious. To keep things reasonably answerable, let me restrict to the following natural candidates (playing fast and loose with isomorphism-type-vs.-group issues for simplicity) - I would love an answer, or even partial progress, to either question (or anything similar really, I'm profoundly stuck-in-the-weeds here):

Q1: Is there an $\mathfrak{A}$ such that $\mathbb{G}(\mathfrak{A})$ consists exactly of the finite $p$-groups for some prime $p$?

Certainly it's possible to get only $2$-groups - see e.g. this MSE answer of Eric Wofsey. However, getting exactly the $2$-groups, or $p$-groups for any fixed $p$, seems much harder.

Q2: Is there a $\mathfrak{B}$ such that $\mathbb{G}(\mathfrak{B})$ consists exactly of the finite abelian groups?

I really have no relevant information for this question, but it seems like a natural one to ask.

Note that Keith Kearnes' answer below addresses a variant of this question, in which variable appearance is replaced by variable dependence, and does not seem to immediately generalize to address this version.

Here are some example $\mathbb{G}(\mathfrak{A})$s (ignoring up-to-isomorphism details):

• If $\mathfrak{A}=(A;\star)$ where $\star:A^2\rightarrow A$ is a bijection, then $\mathbb{G}(\mathfrak{A})$ consists only of the trivial group.

• If $\mathfrak{A}=(\mathbb{N};\max)$, then $\mathbb{G}(\mathfrak{A})=\{S_n:n\in\mathbb{N}\}$.

• Although in the natural numbers with exponentiation alone there is essentially only one nontrivial equation, it turns out that every finite group shows up - see here.

• One of the few negative results I know is that the class of finite cyclic groups is not of the form $\mathbb{G}(\mathfrak{A})$ for any $\mathfrak{A}$. To see this, suppose $t(x_1,...,x_k)$ is a $\mathfrak{A}$-term with $G_\mathfrak{A}(t)\cong C_2$. Then the term $$s(x_{1,1},...,x_{k,k}):=t(t(x_{1,1},...,x_{1,k}), ..., t(x_{k,1},...,x_{k,k}))$$ has $G_\mathfrak{A}(s)$ having too many elements of order $2$ to be cyclic. In fact, tweaking this argument we get that $\mathbb{G}(\mathfrak{A})$ consists entirely of cyclic groups iff $\mathbb{G}(\mathfrak{A})$ consists only of trivial groups. However, this sort of idea doesn't seem to be useful for either Q1 or Q2 above.

Let $\mathfrak{A}$ be an algebra (in the sense of universal algebra). To each term $t(x_1,...,x_n)$ in the language of $\mathfrak{A}$ in which each variable actually appears we can assign a group $G_\mathfrak{A}(t)\subseteq S_n$ consisting of all permutations of the variables which results in the same function: $$G_\mathfrak{A}(t)=\{\sigma\in S_n: \forall a_1,...,a_n\in\mathfrak{A}(t(a_1,...,a_n)=t(a_{\sigma(1)},...,a_{\sigma(n)})\}.$$

Now let $\mathbb{G}(\mathfrak{A})$ be the class of isomorphism types of groups of the form $G_\mathfrak{A}(t)$ for some term $t$. I'm curious which classes of groups can arise this way.

Originally I asked for a complete answer to this question, but in retrospect that was overly ambitious. To keep things reasonably answerable, let me restrict to the following natural candidates (playing fast and loose with isomorphism-type-vs.-group issues for simplicity) - I would love an answer, or even partial progress, to either question (or anything similar really, I'm profoundly stuck-in-the-weeds here):

Q1: Is there an $\mathfrak{A}$ such that $\mathbb{G}(\mathfrak{A})$ consists exactly of the finite $p$-groups for some prime $p$?

Certainly it's possible to get only $2$-groups - see e.g. this MSE answer of Eric Wofsey. However, getting exactly the $2$-groups, or $p$-groups for any fixed $p$, seems much harder.

Q2: Is there a $\mathfrak{B}$ such that $\mathbb{G}(\mathfrak{B})$ consists exactly of the finite abelian groups?

I really have no relevant information for this question, but it seems like a natural one to ask.

Note that Keith Kearnes' answer below addresses a variant of this question, in which variable appearance is replaced by variable dependence, and does not seem to immediately generalize to address this version.

Here are some example $\mathbb{G}(\mathfrak{A})$s (ignoring up-to-isomorphism details):

• If $\mathfrak{A}=(A;\star)$ where $\star:A^2\rightarrow A$ is a bijection, then $\mathbb{G}(\mathfrak{A})$ consists only of the trivial group.

• If $\mathfrak{A}=(\mathbb{N};\max)$, then $\mathbb{G}(\mathfrak{A})=\{S_n:n\in\mathbb{N}\}$.

• Although in the natural numbers with exponentiation alone there is essentially only one nontrivial equation, it turns out that every finite group shows up - see here.

• One of the few negative results I know is that the class of finite cyclic groups is not of the form $\mathbb{G}(\mathfrak{A})$ for any $\mathfrak{A}$. To see this, suppose otherwise. Then thinking about how $C_2$ is represented, we have that $\mathfrak{A}$ has a term $t(x_1,x_2,y_1,...,y_k)$ (with perhaps $k=0$) whose only nontrivial symmetry swaps $x_1$ and $x_2$. But now consider the term $t(t(x_1,x_2,y_1,...,y_k), t(x_3,x_4, z_1,...,z_k), w_1,...,w_k)$. This term's symmetry group contains too many elements of order $2$ to be cyclic. Unfortunately, I don't see how this sort of idea can address either Q1 or Q2 above.

Let $\mathfrak{A}$ be an algebra (in the sense of universal algebra). To each term $t(x_1,...,x_n)$ in the language of $\mathfrak{A}$ in which each variable actually appears we can assign a group $G_\mathfrak{A}(t)\subseteq S_n$ consisting of all permutations of the variables which results in the same function: $$G_\mathfrak{A}(t)=\{\sigma\in S_n: \forall a_1,...,a_n\in\mathfrak{A}(t(a_1,...,a_n)=t(a_{\sigma(1)},...,a_{\sigma(n)})\}.$$

Now let $\mathbb{G}(\mathfrak{A})$ be the class of isomorphism types of groups of the form $G_\mathfrak{A}(t)$ for some term $t$. I'm curious which classes of groups can arise this way.

Originally I asked for a complete answer to this question, but in retrospect that was overly ambitious. To keep things reasonably answerable, let me restrict to the following natural candidates (playing fast and loose with isomorphism-type-vs.-group issues for simplicity) - I would love an answer, or even partial progress, to either question (or anything similar really, I'm profoundly stuck-in-the-weeds here):

Q1: Is there an $\mathfrak{A}$ such that $\mathbb{G}(\mathfrak{A})$ consists exactly of the finite $p$-groups for some prime $p$?

Certainly it's possible to get only $2$-groups - see e.g. this MSE answer of Eric Wofsey. However, getting exactly the $2$-groups, or $p$-groups for any fixed $p$, seems much harder.

Q2: Is there a $\mathfrak{B}$ such that $\mathbb{G}(\mathfrak{B})$ consists exactly of the finite abelian groups?

I really have no relevant information for this question, but it seems like a natural one to ask.

Note that Keith Kearnes' answer below addresses a variant of this question, in which variable appearance is replaced by variable dependence, and does not seem to immediately generalize to address this version.

Here are some example $\mathbb{G}(\mathfrak{A})$s (ignoring up-to-isomorphism details):

• If $\mathfrak{A}=(A;\star)$ where $\star:A^2\rightarrow A$ is a bijection, then $\mathbb{G}(\mathfrak{A})$ consists only of the trivial group.

• If $\mathfrak{A}=(\mathbb{N};\max)$, then $\mathbb{G}(\mathfrak{A})=\{S_n:n\in\mathbb{N}\}$.

• Although in the natural numbers with exponentiation alone there is essentially only one nontrivial equation, it turns out that every finite group shows up - see here.

• One of the few negative results I know is that the class of finite cyclic groups is not of the form $\mathbb{G}(\mathfrak{A})$ for any $\mathfrak{A}$. To see this, suppose $t(x_1,...,x_k)$ is a $\mathfrak{A}$-term with $G_\mathfrak{A}(t)\cong C_2$. Then the term $$s(x_{1,1},...,x_{k,k}):=t(t(x_{1,1},...,x_{1,k}), ..., t(x_{k,1},...,x_{k,k}))$$ has $G_\mathfrak{A}(s)$ having too many elements of order $2$ to be cyclic. However, this sort of idea doesn't seem to be useful for either Q1 or Q2 above.

Let $\mathfrak{A}$ be an algebra (in the sense of universal algebra). To each term $t(x_1,...,x_n)$ in the language of $\mathfrak{A}$ in which each variable actually appears we can assign a group $G_\mathfrak{A}(t)\subseteq S_n$ consisting of all permutations of the variables which results in the same function: $$G_\mathfrak{A}(t)=\{\sigma\in S_n: \forall a_1,...,a_n\in\mathfrak{A}(t(a_1,...,a_n)=t(a_{\sigma(1)},...,a_{\sigma(n)})\}.$$

Now let $\mathbb{G}(\mathfrak{A})$ be the class of isomorphism types of groups of the form $G_\mathfrak{A}(t)$ for some term $t$. I'm curious which classes of groups can arise this way.

Originally I asked for a complete answer to this question, but in retrospect that was overly ambitious. To keep things reasonably answerable, let me restrict to the following natural candidates (playing fast and loose with isomorphism-type-vs.-group issues for simplicity) - I would love an answer, or even partial progress, to either question (or anything similar really, I'm profoundly stuck-in-the-weeds here):

Q1: Is there an $\mathfrak{A}$ such that $\mathbb{G}(\mathfrak{A})$ consists exactly of the finite $p$-groups for some prime $p$?

Certainly it's possible to get only $2$-groups - see e.g. this MSE answer of Eric Wofsey. However, getting exactly the $2$-groups, or $p$-groups for any fixed $p$, seems much harder.

Q2: Is there a $\mathfrak{B}$ such that $\mathbb{G}(\mathfrak{B})$ consists exactly of the finite abelian groups?

I really have no relevant information for this question, but it seems like a natural one to ask.

Note that Keith Kearnes' answer below addresses a variant of this question, in which variable appearance is replaced by variable dependence, and does not seem to immediately generalize to address this version.

Here are some example $\mathbb{G}(\mathfrak{A})$s (ignoring up-to-isomorphism details):

• If $\mathfrak{A}=(A;\star)$ where $\star:A^2\rightarrow A$ is a bijection, then $\mathbb{G}(\mathfrak{A})$ consists only of the trivial group.

• If $\mathfrak{A}=(\mathbb{N};\max)$, then $\mathbb{G}(\mathfrak{A})=\{S_n:n\in\mathbb{N}\}$.

• Although in the natural numbers with exponentiation alone there is essentially only one nontrivial equation, it turns out that every finite group shows up - see here.

Let $\mathfrak{A}$ be an algebra (in the sense of universal algebra). To each term $t(x_1,...,x_n)$ in the language of $\mathfrak{A}$ in which each variable actually appears we can assign a group $G_\mathfrak{A}(t)\subseteq S_n$ consisting of all permutations of the variables which results in the same function: $$G_\mathfrak{A}(t)=\{\sigma\in S_n: \forall a_1,...,a_n\in\mathfrak{A}(t(a_1,...,a_n)=t(a_{\sigma(1)},...,a_{\sigma(n)})\}.$$

Now let $\mathbb{G}(\mathfrak{A})$ be the class of isomorphism types of groups of the form $G_\mathfrak{A}(t)$ for some term $t$. I'm curious which classes of groups can arise this way.

Originally I asked for a complete answer to this question, but in retrospect that was overly ambitious. To keep things reasonably answerable, let me restrict to the following natural candidates (playing fast and loose with isomorphism-type-vs.-group issues for simplicity) - I would love an answer, or even partial progress, to either question (or anything similar really, I'm profoundly stuck-in-the-weeds here):

Q1: Is there an $\mathfrak{A}$ such that $\mathbb{G}(\mathfrak{A})$ consists exactly of the finite $p$-groups for some prime $p$?

Certainly it's possible to get only $2$-groups - see e.g. this MSE answer of Eric Wofsey. However, getting exactly the $2$-groups, or $p$-groups for any fixed $p$, seems much harder.

Q2: Is there a $\mathfrak{B}$ such that $\mathbb{G}(\mathfrak{B})$ consists exactly of the finite abelian groups?

I really have no relevant information for this question, but it seems like a natural one to ask.

Note that Keith Kearnes' answer below addresses a variant of this question, in which variable appearance is replaced by variable dependence, and does not seem to immediately generalize to address this version.

Here are some example $\mathbb{G}(\mathfrak{A})$s (ignoring up-to-isomorphism details):

• If $\mathfrak{A}=(A;\star)$ where $\star:A^2\rightarrow A$ is a bijection, then $\mathbb{G}(\mathfrak{A})$ consists only of the trivial group.

• If $\mathfrak{A}=(\mathbb{N};\max)$, then $\mathbb{G}(\mathfrak{A})=\{S_n:n\in\mathbb{N}\}$.

• Although in the natural numbers with exponentiation alone there is essentially only one nontrivial equation, it turns out that every finite group shows up - see here.

• One of the few negative results I know is that the class of finite cyclic groups is not of the form $\mathbb{G}(\mathfrak{A})$ for any $\mathfrak{A}$. To see this, suppose otherwise. Then thinking about how $C_2$ is represented, we have that $\mathfrak{A}$ has a term $t(x_1,x_2,y_1,...,y_k)$ (with perhaps $k=0$) whose only nontrivial symmetry swaps $x_1$ and $x_2$. But now consider the term $t(t(x_1,x_2,y_1,...,y_k), t(x_3,x_4, z_1,...,z_k), w_1,...,w_k)$. This term's symmetry group contains too many elements of order $2$ to be cyclic. Unfortunately, I don't see how this sort of idea can address either Q1 or Q2 above.