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??

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    $\begingroup$ It's not enough to check that $S_4$ is not isomorphic to $\Bbb Z_6\times\Bbb Z_4$, you need to check also that it is not isomorphic to a semidirect product of these two groups. $\endgroup$ – Stahl Dec 5 '16 at 6:08
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    $\begingroup$ The kernel can't be isomorphic to $\mathbb Z_6$ because $S_4$ doesn't contain an element of order $6$. $\endgroup$ – Bungo Dec 5 '16 at 6:11
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    $\begingroup$ Instead try using the fact that a normal subgroup is a union of conjugacy classes. Then recall that two elements of $S_n$ are conjugate if and only if they have the same cycle type. $\endgroup$ – Ethan Alwaise Dec 5 '16 at 6:20
  • $\begingroup$ Slight variation on the suggestion by @EthanAlwaise : if the kernel $K$ is isomorphic to $S_3$, then it has a unique, hence characteristic, subgroup $H$ of order $3$. Then since $H$ is characteristic in $K$ and $K$ is normal in $S_4$, this means that $H$ is normal in $S_4$. But that can't be true: $H$ must be generated by a $3$-cycle, say $(abc)$, and all such cycles are conjugate. $\endgroup$ – Bungo Dec 5 '16 at 6:28

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?


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