Show that any abelian transitive subgroup of $S_n$ has order $n$ Can anybody tell me what is known about the classification of abelian transitive groups of the symmetric groups?

Let $G$ be a an abelian transitive subgroup of the symmetric group $S_n$. Show that $G$ has order $n$.

Thanks for your help! 
 A: The question is answered by user641 in the comments.


*

*Every subgroup of $S_n$ acts faithfully on $\{1,\cdots,n\}$. This means that no two elements in the subgroup act like the same function on this set.

*A set $X$ on which a group $G$ acts transitively is a single orbit. In particular, it is isomorphic as a $G$-set$^\dagger$ to a coset space $G/H$. Such an isomorphism can be obtained by picking an $x\in X$ and then constructing the correspondence $gx\leftrightarrow g{\rm Stab}_G(x)$ (so, here $H={\rm Stab}_G(x)$).

*If $G$ is abelian, then every element of $H$ acts the same way on $G/H$, so the action of $G$ on the coset space $G/H$ is faithful if and only if $H=1$.


Given our hypotheses, we obtain $\{1,\cdots,n\}\cong^\dagger G/H$ and by the second bullet point, we know the action is faithful by the first bullet point, and therefore we know $H=1$ by the third bullet point; thus we have proved $\{1,\cdots,n\}\cong G/1$, so $|G|=n$.
($^\dagger $A morphism of $G$-sets is a $G$-equivariant aka intertwining map, i.e. a map $\phi:X\to Y$ with the property that $\phi(gx)=g\phi(x)$ for all $x\in X$ and $g\in G$. In fact $G$-sets thus become a category.)
A: Reading this question, an alternative solution came to my mind. It is shorter than my original solution, but slightly less elementary as it uses (very basic) theory of group actions.

By transitivity, the orbit of $1$ is $G\cdot 1 = \{1,\ldots,n\}$.
Let $\sigma\in G_1$, where $G_1 = \{\sigma\in G \mid \sigma(1) = 1\}$ is the stabilizer of $1$.
Let $x\in \{1,\ldots,n\}$. Transitivity gives a $\tau\in G$ with $\tau(1) = x$. Now
$$\sigma(x) = \sigma\tau(1) \overset{G\text{ abelian}}{=} \tau\sigma(1) = \tau(1) = x,$$
showing that $\sigma = \operatorname{id}$ and therefore $G_1 = \{\operatorname{id}\}$.
Now by the orbit-stabilizer theorem,
$$\#G = \#(G\cdot 1) \cdot \#G_1 = n\cdot 1 = n.$$
A: The following solution only needs basic group theory.
Let $G$ be an transitive abelian subgroup of $S_n$.
By transitivity, for each $x\in\{1,\ldots,n\}$ there is a $\sigma\in G$ such that $\sigma(1) = x$. Below, we show that $\sigma$ is uniquely determined by $x$, implying that $\#G = n$.
For showing uniqueness, let $\sigma,\tau\in G$ with $\sigma(1) = \tau(1) = x$. Let $y\in\{1,\ldots,n\}$. From transitivity we get a $\pi\in G$ with $\pi(x) = y$.
Now $$\sigma(y) = \sigma\pi(x) = \sigma\pi\tau(1) \overset{G \text{ abelian}}{=}\tau\pi\sigma(1) = \tau\pi(x) = \tau(y)$$
and therefore $\sigma = \tau$.
