Do any two structures of a language with constants have a common non-trivial substructure? Suppose you have two $\mathcal{L}$-structures $\mathcal{M},\mathcal{N}$. Extend the language to include new distinct constant symbols $c_1,\ldots,c_n$ and take $\mathcal{C}=\langle c_1^{\mathcal{M}},...,c_n^{\mathcal{M}}\rangle$--the substructure generated by the interpretations of the $c_i$ in $\mathcal{M}$. Then $\mathcal{C}\subseteq\mathcal{M}$ and $\mathcal{N}$ contains an isomorphic copy of $\mathcal{C}$. So up to isomorphism, we have a common $\mathcal{L}(c_1,\ldots,c_n)$-structure.
It's a bit unclear to me why I can't then take an $\mathcal{L}$-reduct of $\mathcal{C}$ and deduce that $\mathcal{M}$ and $\mathcal{N}$ have a common non-trivial $\mathcal{L}$-substructure. But the conclusion that any to $\mathcal{L}$-structures have a common non-trivial substructure seems wrong to me.
 A: Welcome to MSE!
Let me expand on my earlier comment, because your intuition is sound: something has gone wrong here.
I think the misunderstanding here has to do with how $\langle c_1^\mathcal{M}, \ldots, c_n^\mathcal{M} \rangle$ is built. It's the smallest substructure of $\mathcal{M}$ containing all the $c_1, \ldots, c_n$. In particular, the elements that wind up in $\langle c_1^\mathcal{M}, \ldots, c_n^\mathcal{M} \rangle$, and the relations between the $c_k^\mathcal{M}$ are heavily dependent on $\mathcal{M}$.
For example, say $\mathcal{M} = C_2$, the cyclic group of order $2$, and $\mathcal{N} = C_3$. Then say that $c^\mathcal{M}$ picks out the generator of $\mathcal{M}$ and $c^\mathcal{N}$ picks out a generator of $\mathcal{N}$. Then $\langle c^\mathcal{M} \rangle = C_2$ (the smallest subgroup containing $c^\mathcal{M}$) and $\langle c^\mathcal{N} \rangle = C_3$ (the smallest subgroup containing $c^\mathcal{N}$).
Of course, these substructures $\langle c^\mathcal{M} \rangle$ and $\langle c^\mathcal{N} \rangle$ are not isomorphic! The issue is that, even though they're both generated by one element, the relations that the elements satisfy are different because they're living inside of different groups. For instance, $(c^\mathcal{M})^2 = e^\mathcal{M}$ while $(c^\mathcal{N})^2 \neq e^\mathcal{N}$. Unfortunately, the notation $\langle c^\mathcal{M} \rangle$ doesn't really indicate this, especially if you leave out the superscripts reminding you what structure you're in.

I hope this helps ^_^
