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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!

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You mean, abelian subgroups of $S_n$ that act transitively on $\{1,\ldots,n\}$? – Arturo Magidin Apr 4 '12 at 19:26
A transitive group action is the same as the coset action on some subgroup. For that action to be faithful, the subgroup in question can contain no normal subgroups. Now think about what G being abelian means. – user641 Apr 4 '12 at 19:29
Anon, How can you show that: Every subgroup of Sn acts faithfully on {1, 2, ..., n}. Thank you. – Thang Dec 26 '14 at 2:40

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.)

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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.

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