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.


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.

  • $\begingroup$ Just a matter of choice: $x\mapsto - x$ is an automorphism of $\Bbb Z_3 /3\Bbb Z$. $\endgroup$ – Berci Apr 21 at 18:26
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    $\begingroup$ wouldn't this cause an issue in the representation table of $A_4$? the columns corresponding to the conjugacy classes of $(123)$ and $(132)$ would be interchangeable for dimension 1 representations? $\endgroup$ – PerelMan Apr 21 at 18:53
  • $\begingroup$ Lining up the relevant representations of $A_4$ with the representations of $\mathbb{Z}_3$ depends on the choice of isomorphism $A_4/V_4\to \mathbb{Z}_3$, not the other way around. $\endgroup$ – David Hill Apr 24 at 17:48
  • $\begingroup$ @DavidHill Then if I choose to assign $(132)V4$ coset to $\overline{1}\in \mathbb{Z}_3$, would the character table of $A_4$ still be correct? $\endgroup$ – PerelMan Apr 24 at 17:58
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    $\begingroup$ The character table does not depend on this assignment. If you assign $(132)V_4$ to $\bar{1}$, then the $1$-d rep of $A_4$ where $V_4$ acts as $1$ and $(123)$ acts as $e^{2\pi i/3}$ corresponds to the representation of $\mathbb{Z}_3$ where $\bar{1}$ acts by $e^{4\pi i/3}$. $\endgroup$ – David Hill Apr 24 at 20:13

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