This question was asked here, but it is not trivial to me and I am stuck in some of the details.
The statement:
let $L$ be a first-order language, and let $M$ and $N$ be two $L$-structures. If $M\equiv N$ ($M$ and $N$ are elementarily equivalent) and $M$ is finite, then $M\cong N$ ($M$ and $N$ are isomorphic).
What I know how to do:
First of all, the domains $U_{M}$ and $U_{N}$ must have the same (finite) cardinality. Indeed, let $n=|U_{M}|$. The $L$-sentence $$\exists\,x_{1}\ldots\exists\,x_{n}\,\left(\bigwedge_{1\leq i<j\leq n}\neg(x_{i}=x_{j})\wedge\forall\,y\left(\bigvee_{1\leq i\leq n}(y=x_{i})\right)\right)$$ is satisfied by $M$, and so it must be by $N$. Therefore, $n=|U_{N}|$ as well.
Second, suppose the result holds for a finite languages, and suppose $M\not\cong N$. Then, each of the $n!$ bijections between $U_{M}$ and $U_{N}$ fails to be an isomorphism. That is, for each such bijection $h$ there is at least one $L$-symbol $l_{h}\in L$ for which $h$ fails to carry its $M$-interpretation to its $N$-interpretation. Choose one $l_{h}$ for each bijection $h$ and collect these $L$-symbols in a set $L^{-}\subseteq L$. Now, since $M\not\cong N$, the reducts $M^{-}=M|_{L^{-}}$ and $N^{-}=N|_{L^{-}}$ are also not isomorphic $L^{-}$-structures. Therefore, $M\not\equiv N$, since the results holds for finite languages. This is a contradiction and $M\cong N$.
What is giving me trouble:
Proving the result for finite languages. Suppose $L$ is finite, that $M\equiv N$ but $M\not\cong N$, and that for each of the $n!$ bijections $h$ between $U_{M}$ and $U_{N}$ we have choosen one $L$-symbol $l_{h}$ where $h$ fails to be an somorphism. The idea seems to be using these symbols somehow to define, for each $h$, an $L$-formula $\varphi_{h}$, and them combine these finitely many $\varphi_{h}$ to define an $L$-sentence $\varphi$ which is satisfied by $M$ but not by $N$, thus reaching a contradiction.
My attempt to formalizing the idea considers the following cases: Let $U_{M}=\{a_{1},\ldots, a_{n}\}$. Let $\alpha$ be the variable assignment $x_{k}\mapsto a_{k}$, $1\leq k\leq n$.
- $l_{h}$ is an $m$-ary predicate $R$: then there are $a_{i_1},\ldots,a_{i_m}\in U_{M}$ such that either $(a_{i_1},\ldots, a_{i_m})\in R^{M}$ and $(h(a_{i_1}),\ldots, h(a_{i_m}))\not\in R^{N}$ OR $(a_{i_1},\ldots, a_{i_m})\not\in R^{M}$ and $(h(a_{i_1}),\ldots, h(a_{i_m}))\in R^{N}$. Let $\varphi_{h}$ be $$Rx_{i_1}\cdots x_{i_m}$$ in the first case and $$\neg Rx_{i_1}\cdots x_{i_m}$$ in the second. Clearly, $M\vDash\varphi_{h}[\alpha]$ and $N\not\vDash\varphi_{h}[h\circ\alpha]$.
- $l_{h}$ is an $m$-ary operation $f$: then there are $a_{k}, a_{i_1},\ldots, a_{i_m}\in U_{M}$ such that $a_{k}=f^{M}(a_{i_1},\ldots,a_{i_m})$ and $h(a_{k})\neq f^{M}(h(a_{i_1}),\ldots h(a_{i_m}))$. Let $\varphi_{h}$ be $$x_{k}=fx_{i_1}\cdots x_{i_m}.$$ Again we get $M\vDash\varphi_{h}[\alpha]$ and $N\not\vDash\varphi_{h}[h\circ\alpha]$.
- $l_{h}$ is a constant $c$: then $h(c^{M})\neq c^{M}$, so there is $a_{k}\in U_{M}$ such that $a_{k}=c^{M}$ and $h(a_{k})\neq c^{N}$. Let $\varphi_{h}$ be $$x_{k}=c,$$ so that $M\vDash\varphi_{h}[\alpha]$ and $N\not\vDash\varphi_{h}[h\circ\alpha]$.
At this point, I am not sure how to combine these $\varphi_{h}$ into and $L$-sentence $\varphi$ satisfied by $M$ but not by $N$. How do I do this? Did I make any mistakes in 1-3?
You've got essentially everything that you need. All that's left is to write the sentence that asserts there are $x_1, \ldots, x_n$, they are different, everything in the universe is one of them, and that each $\phi_h$ holds for the indices you found (since there are only finite many $\phi_h$s, you can do this with a finite conjunction).