Been trying to solve the following logic question:
Show that the class:
$\mathscr G$ = {$G$ is a group| $G\subseteq$ $SL_n$($\mathbb F$), where n $\in$ $\mathbb N$, $\mathbb F$ is any field}
is closed under elementary equivalence (use Keisler–Shelah Isomorphism theorem).
Sketch of the proof (not sure at all if its the correct way) :
Let's consider two groups G$_1$ and G$_2$ in class $\mathscr G$ that are elementarily equivalent. We need to show that there exists an isomorphism between G$_1$ and G$_2$ up to a reduct.
Now, let's construct a new structure H as follows:
The underlying set of H is the union of the underlying sets of G$_1$ and G$_2$.
For each operation (e.g., multiplication, inverse) in the language of groups, define the corresponding operation in H as follows:
If the operation is defined on G$_1$, then use the corresponding operation from G$_1$.
If the operation is defined on G$_2$, then use the corresponding operation from G$_2$.
If the operation is not defined on either G$_1$ or G$_2$, then define it arbitrarily for elements in the intersection of G$_1$ and G$_2$.
It can be shown that the structure H is a group, and by construction, it contains both G$_1$ and G$_2$ as subgroups. Furthermore, since G$_1$ and G$_2$ are elementarily equivalent, the first-order formulas that hold in G$_1$ also hold in G$_2$, and vice versa. Therefore, the first-order sentences satisfied by G$_1$ and G$_2$ are also satisfied by H.
By the Keisler–Shelah Isomorphism theorem, since G$_1$ and G$_2$ are elementarily equivalent, there exists an isomorphism between G$_1$ and G$_2$ up to a reduct. In this case, the reduct is the structure H.
Therefore $\mathscr G$ is closed under elementary equivalence.
Best Answer
I'll give you some pointers about how to get started.
What does it mean to show that $\mathscr{G}$ is closed under elementary equivalence? Well, we need to show that if $G\in \mathscr{G}$ and $G\equiv H$, then $H\in \mathscr{G}$.
Ok, so let $G$ and $H$ be groups such that $G\in \mathscr{G}$ and $H\equiv G$. The fact that $G\in \mathscr{G}$ means that there exists a field $\mathbb{F}$ and $n\in \mathbb{N}$ such that $G\subseteq \mathrm{SL}_n(\mathbb{F})$. And we need to show that there exists a field $\mathbb{K}$ and $m\in \mathbb{N}$ such that $H\subseteq \mathrm{SL}_m(\mathbb{K})$.
[Aside: We need to interpret $H\subseteq \mathrm{SL}_m(\mathbb{K})$ here as "$H$ is isomorphic to a subgroup of $\mathrm{SL}_m(\mathbb{K})$". Indeed, if the elements of $H$ aren't $m\times m$ matrices, we have no hope of showing that $H$ is literally a subset of $\mathrm{SL}_m(\mathbb{K})$. In other words, we should assume that class $\mathscr{G}$ in the question is closed under isomorphism of groups.]
Now the hint says to use the Keisler-Shelah Isomorphism theorem. This theorem tells us that since $G\equiv H$, there exists a set $I$ and an ultrafilter $\mathcal{U}$ on $I$ such that $G^I/\mathcal{U} \cong H^I/\mathcal{U}$. Here $G^I/\mathcal{U}$ is the ultrapower of $G$ by the ultrafilter $\mathcal{U}$.
This is how far you get just by unpacking the definitions, the problem statement, and the hint. I make this comment just because in your attempt in the question, you had already gone quite far astray before this point.
Now how can we possibly use this ultrafilter to produce a field whose special linear group contains $H$? A natural guess would be to consider the ultrapower $\mathbb{F}^I/\mathcal{U}$, which is a field by Łoś's Theorem. This leads us to the question: If $G\subseteq \mathrm{SL}_n(\mathbb{F})$, is $G^I/\mathcal{U}\subseteq \mathrm{SL}_n(\mathbb{F}^I/\mathcal{U})$? Which is really the heart of the problem. I'll note that this is closely related to your previous question.
So to finish: