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 $x\in\{1,\ldots,n\}$ 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 $\sigma = \tau$.

Answer 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.)

Answer 3

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.

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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, $$\#G = \#(G\cdot 1) \cdot \#G_1 = n\cdot 1 = n.$$