subgroup of $GL_n(\mathbb{C})$ which is isomorphic to $S_4$ I want to find the subgroup of $GL_n(\mathbb{C})$ which is isomorphic to $S_4$, $4$th symmetry group. First $n=4$,let $A,B\in GL_4(\mathbb{C})$ be 
$$ A=\left(\begin{array}{cccc}0&0&0&1\\1&0&0&0\\0&1&0&0\\0&0&1&0 \end{array}\right),B=\left(\begin{array}{cccc}0&1&0&0\\1&0&0&0\\0&0&1&0\\0&0&0&1\end{array}\right)$$
Then the subgroup of $GL_4(\mathbb{C})$ generated by $A$ and $B$ is isomorphic to $S_4$. Similarly there are subgroups of $GL_n(\mathbb{C})$ isomorphic to $S_4$ when $n\geq 5$. 
My question: Are there subgroups of $GL_3(\mathbb{C})$ and $GL_2(\mathbb{C})$ which is isomorphic to $S_4$? Please give me some advice. 
 A: Notice that your $4\times 4$ permutation matrices invariante a subspace of $\dim 3$ given by $x_1 + x_2 + x_3 + x_4 = 0$. Taking a basis of this subspace $e_i - e_{i+1}$, $1\le i \le 3$ gives a subgroup of $GL(3, \mathbb{C})$ ( in fact $GL(3, \mathbb{Z})$) that is isomorphic to $S_4$.
Let's show that there does not exist a subgroup of $GL(2, \mathbb{C})$ that is isomorphic to $S_4$. Assume there exists such one, call it $G$. Since $G$ is finite, there exists a positive hermitian form invariant under $G$. Any such form is obtained from the standard one by composing with a map from $GL(2, \mathbb{C})$. Hence, $G$ is contained in a conjugate of $U(2)$. We may assume $G \subset U(2)$. Now, the subgroup 
$$\{ 1, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)\}$$ is in $A_4$, the derived subgroup of $S_4$. Therefore, there exists a subgroup $K$ of $[G,G] \subset [U(2),U(2)]$ isomorphic to $\mathbb{Z}/2\times \mathbb{Z}/2$. Now, we have an abelian subgroup included in $SU(2)$. We know that any commuting family of unitary operators is diagonalisable. Therefore, we may assume that $K$ is contained in the subgroup of diagonal matrices of $SU(2)$. But this one is isomorphic to $S^1$. Now, any finite subgroup of $S^1$ is cyclic. But $K$ is not cyclic. Here we have a contradiction.
We have showed in fact that there is no subgroup of $GL(2, \mathbb{C})$ (equivalently $SU(2)$) isomorphic to $A_4$. Now, there exists a subgroup of $O(3,\mathbb{R})$  isomorphic to $S_4$ ( the isometries of a tetrahedron), so a subgroup of $SO(3,\mathbb{R})$ isomorphic to $A_4$.
We have a cover 
$$1 \to \{ \pm 1\} \to SU(2)\to SO(3,\mathbb{R})\to 1$$
Denote by $\tilde{A}_4$ the preimage of $A_4$. We have a no split cover
$$1 \to \{\pm 1\} \to \tilde{A}_4 \to A_4 \to 1$$
Recall that $SU(2)$ can be seen as the group of quaternions of norm $1$. Then $\tilde{A}_4$ will be the group of Hurwitz quaternions of norm $1$. The preimage $\tilde{K}$ of $K$ is $\{\pm 1, \pm i , \pm j, \pm k\}$.
$\bf{Added:}$ Something that appeared in the proof: for a compact group like $U(n)$ or $SU(n)$ any abelian subgroup is contained in a maximal torus, while for $SO(3, \mathbb{R})$ already this is not true ( $K$ ). 
