One can get arithmetic progressions as truth sets, as in Joel's comment. Pick non-negative integers $a$ and $b$, pick a finite group $G$ which has at least one representation of degree $a$. Then there is a formula expression the statement "the vector space is a $G$-module which is a sum of irreducible representations of degree $a$ and exactly $b$ trivial summands".

Later: For example, the irreps of $G=(\mathbb Z_3\times\mathbb Z_3)\rtimes\mathbb Z_3$ have degree 1 and 3. It is generated by two elements which have cube equal to the identity, and which commute with their commutator. For example, if we want dimensions to be divisible by $3$, we can say:

$(\exists A,B)(A^3=B^3=[A,[A,B]]=[B,[A,B]]=I \wedge \neg(\exists v,\lambda,\mu)(Av=\lambda v\wedge Bv=\mu v))$

(uppercase letters are matrices, lowercase letters are vectors, greek letters are scalars, and commutators are group commutators) A model for this is a $G$ which does not have one-dimensional submodules. This works for other prime values of $3$.

One can get arithmetic progressions as truth sets, as in Joel's comment. Pick non-negative integers $a$ and $b$, pick a finite group $G$ which has at least one representation of degree $a$. Then there is a formula expression the statement "the vector space is a $G$-module which is a sum of irreducible representations of degree $a$ and exactly $b$ trivial summands".

Later: For example, the irreps of $G=(\mathbb Z_3\times\mathbb Z_3)\rtimes\mathbb Z_3$ have degree 1 and 3. It is generated by two elements which have cube equal to the identity, and which commute with their commutator. For example, if we want dimensions to be divisible by $3$, we can say:

$(\exists A,B)(A^3=B^3=[A,[A,B]]=[B,[A,B]]=I \wedge \neg(\exists v,\lambda,\mu)(Av=\lambda v\wedge Bv=\mu v))$

(uppercase letters are matrices, lowercase letters are vectors, greek letters are scalars, and commutators are group commutators) A model for this is a $G$ which does not have one-dimensional submodules. This works for other prime values of $3$.

Later: A vector space $V$ has a structure of $M_n(k)$-module iff $n\mid\dim V$. This can also be written in the language and it is much simpler that the first example!

One can get arithmetic progressions as truth sets, as in Joel's comment. Pick non-negative integers $a$ and $b$, pick a finite group $G$ which has at least one representation of degree $a$. Then there is a formula expression the statement "the vector space is a $G$-module which is a sum of irreducible representations of degree $a$ and exactly $b$ trivial summands".

One can get arithmetic progressions as truth sets, as in Joel's comment. Pick non-negative integers $a$ and $b$, pick a finite group $G$ which has at least one representation of degree $a$. Then there is a formula expression the statement "the vector space is a $G$-module which is a sum of irreducible representations of degree $a$ and exactly $b$ trivial summands".

Later: For example, the irreps of $G=(\mathbb Z_3\times\mathbb Z_3)\rtimes\mathbb Z_3$ have degree 1 and 3. It is generated by two elements which have cube equal to the identity, and which commute with their commutator. For example, if we want dimensions to be divisible by $3$, we can say:

$(\exists A,B)(A^3=B^3=[A,[A,B]]=[B,[A,B]]=I \wedge \neg(\exists v,\lambda,\mu)(Av=\lambda v\wedge Bv=\mu v))$

(uppercase letters are matrices, lowercase letters are vectors, greek letters are scalars, and commutators are group commutators) A model for this is a $G$ which does not have one-dimensional submodules. This works for other prime values of $3$.