In $S_4$ (permutation group of $4$ elements) there is a normal group with order $3$. True or False? This year I'm taking computer engineering science classes and thus, we get a class in discrete algebra. Therefore, I'm familiar with Galois fields for polynomials in code theory, the RSA algorithm, Chinese remainder theorem, a bit of group theory and so on. It's the typical kind of material for cryptography our professor told us. Another remark is that I tried to state this question as clear as possible because it was explicitly requested on this forum.
This semester I came across Sylow's theorem. I thought I understood the theorem until I recently I came across this exercise:
"True or false, give a counterexample or a proof: In the $S_4$-group (permutation group of $4$ elements) there is a normal group with order $3$."
We all know that the order of this group is $4!=24$. Now, $24$ has prime factors $4,3$ and $2$.
Our understanding of Sylow's theorem I learns us that there exist subgroups with order $2$, $2^2$, $2^3$ and $3$. If the given group has only one Sylow-p-subgroup then that this only Sylow-p-subgroup is a normal subgroup.
Then, we can conclude that out of Sylow's theorem II follows that there exist sylow-$p$-subgroups that are non-trivial of order $3$ and $8$ (since these numbers are the highest power of the prime number in de prime factorisation of $24$).
Then we apply the theorem:
$p=8$ and $m=3$: $3 \pmod 8$ does not give $1 \pmod 8$, so there does exist a normal subgroup with the order of $8$.
The other option gives: $p=8$ and $m=3$, thus $8 \pmod 3 =2$ and $2$ does not equal $1 \pmod 3$ so there exists a normal subgroup with order $3$.
Is my answer on the question correct? Any feedback would be greatly appreciated since my colleagues did not have the same results as me on this question.
 A: Two permutations of $S_n$ have the same cycle type if and only if they are conjugate in $S_n$. So, for every $3$-cycle $\sigma\in S_4$, as $\tau$ runs in $S_4$, $\tau\sigma\tau^{-1}$ spans the whole set of the $3$-cycles in $S_4$, whose size is greater than $2$ (and hence there is more than one subgroup of order $3$). Therefore, for every subgroup $H\le S_4$ of order $3$, whose nontrivial elements are then $3$-cycles, necessarily $\tau H\tau^{-1}\notin H$ for some $\tau\in S_4$.
A: To solve this problem, we have to apply Sylow's Second and Third Theorems.
Let $p$ be a prime. The Sylow's Second Theorem states that all Sylow $p$-subgroups are conjugate. So if a Sylow $p$-subgroup is normal, then it must be unique.
Let $G$ be a group such that $|G|=p^nm$ where $p\not\mid m$ and let $n_p$ be the number of Sylow $p$-subgroups. The Sylow's Third Theorem states that $n_p\equiv 1 \bmod p$ and $n_p\mid m$.
A subgroup of $S_4$ of order $3$ is a Sylow $3$-subgroup. So we first find the number of Sylow $3$-subgroups, $n_3$. Write $|S_4|=24=2^3\cdot3$. By Sylow's Third Theorem, $n_3\equiv 1 \bmod 3$ and $n_3 \mid 4$. The possibilities of $n_3$ are $1$ and $4$. But it can be see that $\langle (123)\rangle$ and $\langle (234)\rangle$ are two distinct Sylow $3$-subgroups. This implies that $n_3$ must be $4$.
Since a Sylow $3$-subgroup is not unique, we conclude by Sylow's Second Theorem that a Sylow $3$-subgroup cannot be normal in $S_4$.
A: In $S_4$ there are $8$ elements of order $3$, each of them laying in one and only one subgroup of order $3$. Therefore, there are $4$ subgroups of order $3$, overall, and hence the claim is false.
