Let $\mathbb{M}=\langle M,\circ, e\rangle$ be a monoid. We say that $\mathbb{M}$ is finitely generated when there exist finitely many elements in $M$ such that each element of $M$ may be generated using only these elements.
For example, $\langle A^*,\circ, \epsilon \rangle $ is finitely generated if only and only alphabet $A$ is finite. Show that the class of finitely generated monoids is not axiomatized.
My solution Let assume that there exists a such set of sentences $\Delta$ that $\mathbb{M} \models \Delta \iff \mathbb{M} \in \mathbb{M_K}$ where $ \mathbb{M_K}$ is a class of finitely generated monoids.
So, let's consider a language $L = \langle A^*, \circ, e \rangle$ where $A = \{a \}$ So $A$ is a finite set, therefore $L$ is a finitely generated monoid. So, $L \models \Delta$.
Let $\Delta' = \Delta \cup \Gamma, \Gamma = \{\exists w w = b_0, \exists w w = b_0 \circ b_1, \exists w w = b_0 \circ b_1 \circ .. \circ b_n | b_0, .., b_n \text { are pairwise different and } n \in \mathbb{N}$.
Now, for every finite $\Delta_0 \subseteq \Delta'$ it is easy to point the such finitely generated $L_2$ that $L_2 \models \Delta_0$. From the compactness theorem there exists the model ( a language $L'$) for $\Delta', L' \models \Delta'$. But, $L' \models \Delta \cup \Gamma $ and it is not possible because $L'$ is not finitely generated while $\Delta$ force that. Contradiction.
Ok?
I would like to solve it using Ehrenfeucht-Fraisse game, but I don't know how to start. Please help.
Especially, what does it mean that two words are in relationship?