[Math] Show that any abelian transitive subgroup of $S_n$ has order $n$

Question

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!

Answer 1

The following solution only needs basic group theory.

Let $G$ be an transitive abelian subgroup of $S_n$. By transitivity, for each $i\in\{1,\ldots,n\}$ there is a $\sigma\in G$ such that $\sigma(1) = i$. So $\# G\geq n$.

Assume that $\#G > n$. Then there are $\sigma, \tau\in G$ with $x := \sigma(1) = \tau(1)$ and $\sigma\neq \tau$. By the second condition, there is a $y\in\{1,\ldots,n\}$ with $\sigma(y) \neq \tau(y)$. From transitivity we get a $\pi\in G$ with $\pi(x) = y$.

Now $$ \pi\tau\pi\sigma(1) = \pi\tau\pi(x) = \pi\tau(y) $$ and $$ \pi\sigma\pi\tau(1) = \pi\sigma\pi(x) = \pi\sigma(y)\text{.} $$ Because of $\tau(y) \neq \sigma(y)$, these two elements are distinct. So the elements $\pi\tau\in G$ and $\pi\sigma\in G$ do not commute, which contradicts the precondition that $G$ is abelian.