Show some $\mathbb{X} \subseteq \mathbb{N}^+$ occurs as the finite spectrum of a sentence for this language.

Question

Setting

Define the finite spectrum of an $\mathcal{L}$-sentence $\phi$ as

$$\{ n \in \mathbb{N}^+ ~:~ there ~ is~ \mathcal{M} \models \phi ~with~ |\mathbb{M}| = n\}$$

And let $$\mathbb{X} = \{ 2^n 3^m ~:~ n,m > 0\}$$

Now for this $\mathbb{X} \subseteq \mathbb{N}^+$, I would like to show $\mathbb{X}$ occurs as the finite spectrum of some $\phi$ for $\mathcal{L}$.

Updated Problem

I am a bit lost at how to find such sentence $\phi$ and language $\mathcal{L}$.

Answer 1

In a language of groups $\mathcal L=\{\cdot, ^{-1},e\}$ take $\phi$ to be the conjunction of:

1. $\cdot$ is associative;

2. $e$ is neutral for $\cdot$;

3. $x^{-1}$ is inverse of $x$;

4. $\forall x(x\cdot x\cdot x\cdot x\cdot x\cdot x=e)$.

If $\mathbb M\models\phi$ is finite model, then 1, 2. and 3. imply that $\mathbb M$ is a group. 4. implies that every element is of order $1$, $2$, $3$, or $6$, hence by Cauchy's lemma $2$ and $3$ are only primes dividing $|\mathbb M|$. Therefore $|\mathbb M|=2^n3^m$, for some $n,m\geq 0$.

On the other hand, for every $n,m\geq 0$ obviously $\mathbb Z_2^n\times \mathbb Z_3^m\models \phi$.

Therefore, the finite spectrum of $\phi$ is $\{2^n3^m\mid n,m\geq 0\}$.