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 $A_4$.
Best Answer
If $\varphi: G \to H$ is a group homomorphism, then given any element $a \in G$, we know that $\text{ord}(a)$ is a multiple of $\text{ord}(f(a))$.
Now we recall little background on the symmetric group $S_4$ and its subgroup $A_4$. The group $S_4$ has elements of cycles: 1-cycle, 2-cycle, 3-cycle, 4-cycles and product of two 2-cycles. Thus the possible orders of the elements of $S_4$ are $1,2,3,4$. But it is a fact that $A_4$ don't have an element of order $4$. So the possible orders of elements of $A_4$ are $1,2,3$.
If there is a surjective homomorphism $\varphi: A_4 \to \mathbb{Z}_4=\{\bar 0, \bar 1, \bar 2, \bar 3\}$. There exists some $a \in A_4$ such that $\varphi(a)=\bar 3$. Thus by the first sentence above, $\text{ord}(a)$ would be a multiple of $\text{ord}(\varphi(a))=\text{ord}(\bar 3)=4$, which is not possible, since $a \in A_4$ can not have order $4$. Thus there is no such surjective homomorphism.