Proving that $A_4/V_4$ can be generated by $\sigma V_4$ where $\sigma$ is a 3-cycle

Question

Define $V_4 = \{ \text{id}, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \}$

I want to prove that given a 3-cycle $\sigma$, $A_4/V_4 = \langle\sigma V_4\rangle$.

So far I have proved that $V_4$ is a normal subgroup of $A_4$ but I'm not sure where to go from there. Any help would be appreciated.

Answer 1

You have already listed the 4 elements of $V_4$ an order 4 subgroup in $A_4$ of order 12, hence index (= number of cosets) is 3. SO two more cosets remain. Write out the cosets containing each of them

That is elements of the form $\sigma\circ \tau$ with $\tau$ selected from $V_4$. and $\sigma^2\circ \tau$. These 3 together will give $4+4+4=12$ elements exhausting $A_4$. Verify that all elements of $A_4$ appear this way and you are done.