Why is $S_3$ not normal in $S_4$?

Question

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.

See also this duplicate. – Dietrich Burde Jun 05 '19 at 19:28 — Dietrich Burde, Jun 05 '19 at 19:28

score 4 · Answer 1 · answered Jun 04 '19 at 23:21

4

Compute $(1,4)(1,2,3)(4,1)$ see that it is not is $S_3$ even though $(1,2,3)$ is.

answered Jun 04 '19 at 23:21

Ned

4,012

score 3 · Answer 2 · answered Jun 04 '19 at 23:24

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 $(123)(4)$.

score 2 · Accepted Answer · answered Jun 04 '19 at 23:18

2

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 $4$.

answered Jun 04 '19 at 23:18

Ethan Bolker

103,433

score 1 · Answer 4 · answered Jun 04 '19 at 23:30

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.

Why is $S_3$ not normal in $S_4$?

4 Answers4