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Over a field, the answer to the first question is no. An element of $A$ of order 4 in SL(2,R) must satisfy the equation $x^{4}-1$. But it also satisfies the characteristic polynomial $x^{2}-{\rm Tr}(A) x+{\rm det}(A)$x+1$. Taking the gcd of these two polynomials, we quickly see that$A$also satisfies${\rm Tr}(A)^{3}x+(1+{\rm Tr}(A)^{2}{\rm det}(A)-{\rm det}(A)^{2})$({\rm Tr}(A)^{3}-2{\rm Tr}(A)x+{\rm Tr}(A)^{2}$. If the coefficient of ${\rm Tr}(A)\neq 0$ x$is nonzero then$A$satisfies a linear polynomial, so is diagonal, thus has 4th roots of unity along the diagonal. There are only finitely many such matrices. If${\rm Tr}(A)=0$the coefficient of that linear polynomial is zero then${\rm det}(A)^{2}=1$, which says that Tr}(A)=0$ and so $A$ is a root of $x^{2}+1$, which means any two elements of order 4 have equal squares (except for the finitely many possibilities expressed in the previous case).
As for the original question: $S_{4}$ has a presentation $\langle s,t| s^{2}=t^{3}=(st)^{4}=1\rangle$. I would suggest looking at the ring $R=\mathbb{Z}[a,b,c,d,e,f,g,h]/I$ where $I$ is the set of relations forcing the 2x2 matrices give by s=((a,b),(c,d)) and t=((e,f),(g,h)) to satisfy the relations for $S_4$. S_4$and have determinant 1. Finding a Grobner-type basis for$I$should demonstrate that the resulting structure has at least 24 distinct matrices in the group generated by s and t (or it doesn't and you cannot embed$S_{4}$in such a ring). 1 Over a field, the answer to the first question is no. An element of$A$of order 4 in SL(2,R) must satisfy the equation$x^{4}-1$. But it also satisfies the characteristic polynomial$x^{2}-{\rm Tr}(A) x+{\rm det}(A)$. Taking the gcd of these two polynomials, we quickly see that$A$also satisfies${\rm Tr}(A)^{3}x+(1+{\rm Tr}(A)^{2}{\rm det}(A)-{\rm det}(A)^{2})$. If${\rm Tr}(A)\neq 0$then$A$satisfies a linear polynomial, so is diagonal, thus has 4th roots of unity along the diagonal. There are only finitely many such matrices. If${\rm Tr}(A)=0$then${\rm det}(A)^{2}=1$, which says that$A$is a root of$x^{2}+1$, which means any two elements of order 4 have equal squares (except for the finitely many possibilities expressed in the previous case). As for the original question:$S_{4}$has a presentation$\langle s,t| s^{2}=t^{3}=(st)^{4}=1\rangle$. I would suggest looking at the ring$R=\mathbb{Z}[a,b,c,d,e,f,g,h]/I$where$I$is the set of relations forcing the 2x2 matrices give by s=((a,b),(c,d)) and t=((e,f),(g,h)) to satisfy the relations for$S_4$. Finding a Grobner-type basis for$I$should demonstrate that the resulting structure has at least 24 distinct matrices in the group generated by s and t (or it doesn't and you cannot embed$S_{4}\$ in such a ring).