# Are normal subgroups of $S_4$ transitive?

Let $N$ be a normal subgroup of $S_4$.

I know there are two nontrivial proper normal subgroups of $S_4$, but let's not assume that.

I have proven that only subgroups of order $4$ which satisfy "for every $i\neq j$, there exists $\sigma \in N$ such that $\sigma(i)=j$" are the 3 cyclic groups and $V=\{id,(12)(34),(14)(23),(13)(24)\}$.

With this information, how do I prove that $V$ is the one and only normal subgroup of $S_4$ whose order is $4$?

• A group of order $4$ is either cyclic or the product of two cyclic groups of order $2$. Because you know very well the order of the elements of $S_4$, you can use this remark to classify completely the 4-element subgroups of $S_4$ and check that $V_4$ is indeed the only normal one. That's really easy, but not very conceptual. – PseudoNeo Jun 4 '15 at 10:59

Normal (non-trivial) subgroups of any $S_n$ must be transitive.
Fix $n$, and take $S_n$ with a non-trivial non-transitive subgroup $H$. Then $H$ divides the $n$ subjects into more than one orbit (because of non-transitivity), and at least one of those orbits are larger than $1$ (because of non-triviality).
Pick one of those larger-than-one orbits, a subject $a$ in that orbit and a $\sigma \in H$ that moves $a$. Now pick a $b$ in any other orbit. I claim that $(ab)H(ab)^{-1} = (ab)H(ab) \neq H$. Indeed, $(ab)\sigma(ab)$ moves $b$ out of its $H$-orbit, then moves it around within the $H$-orbit of $a$, and then fails to put it back in its own $H$-orbit, and therefore that permutation cannot be in $H$, so $H$ is not normal.