Existence of an onto group homomorphism from $S_4$ to $\Bbb Z_4$ 
Let $S_n$ be the symmetric group of $n$ letters. Then does there exist an onto group homomorphism
   from $S_4$ to $\Bbb Z_4$?

My try: Suppose that  $f:S_4 \to \Bbb Z_4$ is a group homommorphism. Then $S_4/\ker f\cong \Bbb Z_4\implies o(\ker f)=6\implies \ker f$ is isomorphic to $S_3$ or $\Bbb Z_6$.
If $\ker f=\Bbb Z_6\implies S_4\cong  \Bbb Z_6\times \Bbb Z_4$ which is false as $S_4$ is not commutative whereas $\Bbb Z_6\times \Bbb Z_4$ is.
If $\ker f=S_3\implies S_3$ is a normal subgroup of $S_4$.
Now take $S_3=\{e,(12),(23),(13),(123),(132)\}$.Then $(14)(123)(14)=(234)\notin S_3$.Hence $S_3$ is not normal.
Is my solution correct??

Your argument up to ''$\ker f=S_3\text{ or }\mathbb{Z}_6$'' is correct.
But after this, it is possible but lengthy to continue the arguments; for example if kernel is isomorphic to $S_3$ then you have taken it equal to $\{(1), (123),..\}$; this is correct but needs a justification.
Better is the following: $|\ker f|=6$, so $\ker f$ contains an element of order $3$. Since elements of order $3$ in $S_4$ are precisely $3$-cycles (easy to prove) and any two $3$-cycles are conjugate, hence all the $3$-cycles of $S_4$ should be in the kernel (since kernel is normal).
But now we get a contradiction. How many $3$-cycles are there in $S_4$? What is size of kernel?
