Why is $S_3$ not normal in $S_4$? I read on this site that $S_3$ is not normal in $S_4$, but I have not been able to prove it. Admittedly, I have not tried brute force because there must be a smart approach. I have tried to find a counterexample where $\alpha\sigma\neq\sigma'\alpha$, where $\sigma$ and $\sigma'$ are permutations that fix the fourth element, but thus far unsuccessfully. Any hints? I don't really know which facts about the group (or group theory in general) to use.
 A: Compute $(1,4)(1,2,3)(4,1)$ see that it is not is $S_3$ even though $(1,2,3)$ is.
A: An example is $(14)(123)(14) = (234)(1) \notin S_3$.
The reasoning behind this comes from the fact that the conjugacy classes of $S_n$ are determined by cycle structures. If we had $S_3$ as a normal subgroup of $S_4$, then all cycles with some cycle type would be contained in $S_3$. This is clearly not true, as $(234)(1)$ can be decomposed into a cycle of length 3 and of length 1, meaning it is in the same conjugacy class as $(123)(4)$.
A: Write down some explicit permutation $\sigma$ that fixes $4$ and then calculate the conjugate $\tau \sigma \tau^{-1}$ for various two-cycles $\tau$ until you find one that produces a permutation that does not fix $4$.
A: There are four copies of $S_3$ in $S_4$, corresponding to permutations that fix one element: $S_3^k = \{ \sigma \in S_4 : \sigma(k)=k \}$. The one we know as $S_3$ is $S_3^4$ in this notation.
These four copies are conjugate: $S_3^k = (k4)S_3^4(k4)$.
Therefore, $S_3^4$ is not normal.
