Infinite dimensional linear groups. Suppose we have a group of matrices over finite field $\mathbb F_p$ such that in every row and in every column there are only finitely many elements from $\mathbb F_p$. (Also every finite set of columns is independent, and we have such condition for rows also.) Is it true that an isomorphism of such group of matrices ($GL(\mathbb F_p)\simeq GL(\mathbb F_q)$) implies the isomorphism of their stable infinite dimensional linear groups ($GL_{\infty}(\mathbb F_p)\simeq GL_{\infty}(\mathbb F_q)$), i.e. those  invertible infinite matrices which differ from the identity matrix in only finitely many places?
NB. We assume $p,q$ -- primes.
Many many thanks in advance for any help! 
 A: Let $R$ be an arbitrary associative commutative ring. Fix an index set $J$, say infinite countable.
We consider matrices indexed by $J\times J$, hence functions $J\times J\to R$. Let $I$ be the identity matrix. Define $M^2(R)$ as the set of matrices such that every column has finitely nonzero many entries. Then matrix multiplication makes sense on $M^2(R)$, making it an $R$-algebra. Let $M^1(R)$ be the subalgebra of matrices for which each row also has finitely many nonzero entries, and define $M^0(K)$ as those matrices $f$ such that there exists $r=u(f)\in R$ such that $f-rI$ has only finitely many nonzero entries (thus $u$ is a $R$-algebra homomorphism $M^0(R)\to R$).
Let $\mathrm{GL}^i(R)$ be the group of multiplicative units in $M^i(R)$. Hence $u$ defines a homomorphism from $\mathrm{GL}^0(R)$ to the group of units $R^\times$; denote by $\mathrm{GL}_\infty(R)$ its kernel, the usual stable general linear group. The latter has a well-defined determinant homomorphism, valued in $R^\times$, its kernel is by definition $\mathrm{SL}_\infty(R)$. Note that $\mathrm{GL}_\infty(R)$ is the set of elements in $\mathrm{GL}^2(R)$ that differ from the identity matrix only at finitely many columns.
Then $\mathrm{GL}_\infty(R)$ is normal in $\mathrm{GL}^1(R)$: indeed fix $g\in \mathrm{GL}^1(R)$ and $f\in\mathrm{GL}_\infty(R)$. Fix a cofinite subset $J'$ of $J$ such that $f(e_i)=e_i$ for all $i\in J'$. By the assumption on rows, there exists $J''$ cofinite such that $g$ maps every $e_i$ for $i\in J''$ into $R^{J'}$. It follows that $g^{-1}fg$ maps every $e_i$ for $i\in J''$ to itself; thus $g^{-1}fg\in \mathrm{GL}_\infty(R)$.
 The centralizer of $\mathrm{SL}_\infty(R)$ in $M^2(R)$ is reduced to the scalar matrices, by a simple argument. In particular, every nontrivial normal subgroup of $\mathrm{GL}^1(R)$ intersects $\mathrm{GL}_\infty(R)$ nontrivially.
Now assume that $R=K$ is a field: then standard arguments show that noncentral normal subgroups of $\mathrm{GL}_\infty(K)$ contain $\mathrm{SL}_\infty(K)$; in particular the latter is characteristic in $\mathrm{GL}_\infty(K)$ and hence is normal in $\mathrm{GL}^1(R)$; moreover the argument shows that $\mathrm{SL}_\infty(K)$ is the unique minimal non-central normal subgroup of $\mathrm{GL}^1(K)$. 
This already shows that the isomorphism type of $\mathrm{SL}_\infty(K)$ is determined by the isomorphism type of $\mathrm{GL}^1(K)$. 
I expect that $\mathrm{GL}_\infty(K)$ can also be determined but it sounds a little more complicated. Since it is the quotient of $K^*\mathrm{GL}_\infty(K)$ by its center, it is enough to show that $K^*\mathrm{GL}_\infty(K)$ can be determined: I expect that it is characterized by the fact that its quotient by $\mathrm{SL}_\infty(K)$ is the unique maximal abelian normal subgroup of $\mathrm{GL}^1(K)/\mathrm{SL}_\infty(K)$. For instance this would be true if $\mathrm{GL}^1(K)/(K^*\mathrm{GL}_\infty(K))$ was shown to be simple, but this sounds even more complicated.
