I am given the question like this:
Let $N = \{a \in S_4 : a(4) = 4\}$. Explain why $N$ is a subgroup of $S_4$ that is isomorphic to $S_3$. Is $N$ normal in $S_4$? Generalize.
My answer:
Since we already fix $a(4) = 4$, so $N$ is basically isomorphic to group of permutations of 3 symbols which is $S_3$.
For the second question, giving counter example: $(14)(12)(41) = (1)(24) \not \in N$ where $(14)$ and $(14)^{-1} = (41) \in S_4$ and $(12) \in N$, therefore, $N$ is not normal in $S_4$.
Can someone please verify if this answer is correct? Thanks a lot.
You also have that $N$ cannot be normal in $S_4$ because a non-trivial normal subgroup of a primitive group is transitive.