# The coset corresponding to permutation $(123)$ in $\Bbb Z /3\Bbb Z$.

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?

## 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 $$\pi((132))$$.

• Just a matter of choice: $x\mapsto - x$ is an automorphism of $\Bbb Z_3 /3\Bbb Z$. – Berci Apr 21 at 18:26
• 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? – PerelMan Apr 21 at 18:53
• 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. – David Hill Apr 24 at 17:48
• @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? – PerelMan Apr 24 at 17:58
• 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 $e^{4\pi i/3}$. – David Hill Apr 24 at 20:13