Consider the following matrices $A:=$$ \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} ,\ B: =\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ of multiplicative group $GL_2(\mathbb{C})$ and $\sigma = (1234), \rho = (24) \in S_4.$ Let $G_1 = \left \langle {A,B}\right \rangle$ and $G_2= \left \langle {\sigma,\rho}\right \rangle.$ Is $G_1 \cong G_2?$
Here is my proof. But I have trouble showing $G_1 \cong G_2.$ Could someone advise please? Thank you.
Proof:
By direct computation, $o(A)=4$ and $o(B)=2.$ Since $A^i \neq B \ (i=0,1,2,3),$ $G_1 = \{A^{i}B^{j}| i=0,1,2,3 \wedge j=0,1\},$ whence $|G_1|=8.$
$\sigma^2=(13)(24), \ \sigma^3=(1432), \ \sigma^4= (1).$ So $o(\sigma)=4.$ Similarly, $o(\rho)=2.$ By similar reasoning above, $G_2= \{\sigma^i\rho^j| i =0,1,2,3 \wedge j=0,1\},$ whence $|G_2|=8.$
Given $A^iB^j \in G_1, $ let $A^iB^j \mapsto \sigma^i\rho^j.$ Clearly, this mapping is surjective. But how do I show this mapping is injective homomorphism?