Does there exist a surjective group homomorphism $\varphi:A_4\rightarrow\mathbb{Z}/4\mathbb{Z}$?
Edit: I have narrowed it down to the problem of whether $A_4$ has a normal subgroup of order 3 with 3-cycles as elements:
From the theorem of group homomorphism if $\varphi$ is a homomorphism then the $\operatorname{ker}(\varphi)$ has to be normal. From the isomorphism theorem the quotient group $A_{4} / \operatorname{ker}(\varphi)$ will be isomorphic to $\mathbb{Z} / 4 \mathbb{Z}$ (If the kernel is normal). Since $\mathbb{Z} / 4 \mathbb{Z}$ has order 4, we know that $A_{4} / \operatorname{ker}(\varphi)$ also has order 4 (since they're isomorphic). From lagranges theorem we can then conclude that $|\operatorname{ker} \phi|=\frac{\left|A_{n}\right|}{\left|A_{4} / \operatorname{ker} \phi\right|}=\frac{12}{4}=3$.
Since 3 is a prime and the order of a group element divides the order of the group, we know that the elements of the kernel must have order 3 (Not order 1, since all 1 cycles are the identity element). So the kernel has to be the group consisting of three 3-cycles from $A_4$.