Suppose $G$ is a abelian, transitive subgroup of $S_n$. Prove that for all $x\in\{1,\dots,n\}$, $G_x=\langle 1\rangle$

Question

Question: Suppose $G$ is a abelian, transitive subgroup of $S_n$. Prove that for all $x\in\{1,\dots,n\}$, $G_x=\langle 1\rangle$, where $G_x$ is the stabilizer of $x$ in $G$.

Attempt: Suppose $1\neq g\in G$ and suppose $xg=x$ for some $x\in\{1,\dots,n\}$. Since $G$ is a transitive subgroup of $S_n$, there is an $h\in G$ such that $x(h)=x_0\in\{1,\dots, n\}$. Then,

$$\begin{equation*} \begin{split} (x_0)g & = ((x)h)g, \text{ by above equality} \\ & = ((x)g)(h)\\ & = h(x) \\ & = x_0 \end{split} \end{equation*}$$ Thus $g=1$, a contradiction. Hence, $G_x=\langle 1 \rangle$, as wanted.

Answer 1

Your solution looks fine. You could polish it a little bit by doing a direct proof, without setting up a contradiction.

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Let $x\in \{1,\ldots,n\}$ and $\sigma\in G_x$. Let $y\in\{1,\ldots,n\}$. From transitivity we get a $\pi\in G$ with $\pi(x) = y$. As $G$ is abelian, $\sigma\pi = \pi\sigma$, so $$ \sigma(y) = \sigma\pi(x) = \pi\sigma(x) = \pi(x) = y\text{.} $$ Since this is true for any $y\in\{1,\ldots,n\}$, we get that $\sigma = \operatorname{id}$.

Answer 2

$$\forall(g,h)\in G\times G_x\;ghg^{-1}gx=gx$$ $\begin{align} &\therefore\;\forall g\in G\; G_{x}=g\cdot G_x\cdot g^{-1}\subseteq G_{gx}\\ &\therefore\;\forall g\in G\;\;G_{gx}\subseteq G_{g^{-1}(gx)}=G_x\\ &\therefore\;\forall g\in G\;G_x=G_{gx}\\ &\therefore\;G_x=\bigcap_{y\in\{1,\dots,n\}}G_y \end{align}$

Therefore, $G_x=\{e\}$ because each $G_x$ member fixes each $\{1,\dots,n\}$ member.