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!

Solution 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.

Solution 2:

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.)