We know that $V_4 \triangleleft A_4$ so $ A_4/V_4 \cong \Bbb Z /3\Bbb Z$.
The coset corresponding to permutation $(123)$ is $(123)V_4$. Is it corresponding to $\overline{1}$ or $\overline{2}$ in $\Bbb Z /3\Bbb Z$?
Both $\overline{1}$ and $\overline{2}$ are of order 3 so I don't see if there is a canonical morphism $ A_4/V_4 \cong \Bbb Z /3\Bbb Z$ or just a matter of choice?
Thank you for your help.
Edit
Let's fix $j=e^{2i\pi/3}$.
Let $(\Bbb C^*, \rho)$ be a 1-dim representation of $A_4$ and let $\pi: A_4\to A_4/V_4$ and $\rho': A_4/V_4\to \Bbb C^*$ so we have $\chi_{\omega}(g)=\rho(g)=\rho'\circ\pi(g)$ for $g\in A_4$.
We know that $\rho'(\overline{k}) = \omega^k$ where $\omega$ is $3$-th root of unity.
If we choose $\pi((123)) = \overline{1}$ then we should have $\omega = j\Rightarrow \chi_j((123))=j, \chi_{j^2}((123))=j^2$
If we choose $\pi((123)) = \overline{2}$ then we should have $\omega = j^2\Rightarrow \chi_j((123))=j^2, \chi_{j^2}((123))=j$ and we don't preserve the character table of $A_4$:
$ \begin{array}{|c|c|c|c|} \hline A_4& id & (12)(34) & (123) & (132)\\ \hline \chi_{id}&1 &1 &1 & 1\\ \hline \chi_{j}& 1 &1 &j & j^2\\ \hline \chi_{j^2}&1 & 1& j^2 & j\\ \hline. \end{array}$
I don't understand why $\omega$ depends on the choice of $\pi((132))$.
Thank you for your help.