How to prove that $S_4$ a symmetric group has transitive action over $X=\{1, 2, 3, 4\}$?

Question

I have this definitions in hands:

Definition 1: Let G be a group and $x$ an element (fixed!) of G. Given $h \in G$ the element $hxh^{-1}$ is called conjugated of $x$ through $h$.

Definition 2: The conjugates of a group G is the set $X=\{hxh^{-1}: h \in G\}$.

Definition 3: The group G acts transitively over a set of conjugates of $x$ taking $(g, hxh^{-1})$ to $ghxh^{-1}g^{-1}$. So, this is a left action of $G$ over $X$.

Definition 4: In this case the stabilizer of $x$ is $G_x=\{g \in G: gxg^{-1}=x\}=\{g \in G: gx=xg\}$ called the centralizer of $x$, written $C_G(x)$.

Answer 1

Your definitions involve the conjugation action. But it is natural to have $S_n$ act on $\{1,2,\dots,n\}$ just by $\pi\cdot x=\pi(x)$.

Now both myself and @MikeEarnest have pointed out that it's easy to see that this action is transitive.

Finally, you don't need the group to be abelian in order for it to act on a set.

Answer 2

My answer:

I know that $S_4=\{1, (12), (13), (14), (23), (24), (34), (123), (132), (134), (143), (124), (142), (234), (243), (12)(34), (13)(24), (14)(23), (1234), (1243), (1324), (1342), (1423), (1432)\}$

My way to solve this question it was doing:

• $1 \in C_G(x)$ because 1 comutes with everybody.

From here, I've done this (to cycles with has same length, because I also know this theorem: two permutations conjugate iff they have the same cycle structure.)

FIRST I FIXED THE ELEMENT $(12)$ Then because of the definition 4, above:

• $(12)(13)=(13)(12)?$. NO, because in the left side of the equality we have: $gf1=g3$, $gf2=g2$, $gf3=g1$, $g3=3$, $g2=1$, $g1=2$. Then $1\mapsto 2$, $2\mapsto 1$ and $3\mapsto 2$. Which is the same of $(12)(13)=(132)$. And, similarly, in the right side of the equality we have: $gf1=g2$, $gf2=g1$, $gf3=g3$, $g2=2$, $g1=3$, $g3=1$. Then $1\mapsto 2$, $2\mapsto 3$ and $3\mapsto 1$. Which is the same of $(12)(13)=(123)$.

However $(123)\neq (132)$.

So, I would like to know how is the smartest way to show that $S_4$ has transitive action over $X=\{1, 2, 3, 4\}?$

Answer 3

Let $i,j∈X:=\{1,\dots,n\}$ be two elements. Define $σ\colon X\to X$ by $σ(i)=j, σ(j)=i$, and $σ(k)=k$ for all $k∈X\setminus\{i,j\}$. Then $σ∈S_n$ because it's a bijection, and it shows that $i$ and $j$ are in the same orbit of the natural action $\tau\cdot k:=\tau(k)$. But $i$ and $j$ are arbitrary, so there is one orbit only for this action, namely $O(i)=X, \forall i∈X$.

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EDIT.

Yet another way to show that the action $\sigma\cdot i:=\sigma(i)$ is transitive is the following.

The stabilizer if $i$ with respect to this action is:

\begin{alignat}{1} \operatorname{Stab}(i)&=\{\sigma\in S_n\mid\sigma\cdot i=i\} \\ &=\{\sigma\in S_n\mid\sigma(i)=i\} \end{alignat}

whence $|\operatorname{Stab}(i)|=(n-1)!, \forall i \in \{1,\dots,n\}$. By the Orbit-Stabilizer Theorem, it follows that $|O(i)|=n!/(n-1)!=n=|\{1,\dots,n\}|, \forall i \in \{1,\dots,n\}$, and thence there is one orbit, only.