Exercise 4.1.3 (Abelian transitive subgroups of $S_A$)

Question

Assume that $G$ is an abelian, transitive subgroup of $S_A$. Show that $\sigma(a) 
e a$ for all $\sigma \in G - \{1\}$ and all $a \in A$. Deduce that $|G| = |A|$. [Use the preceding exercise.]

Richard Ganaye · Accepted Answer

\begin{proof} 
By Exercise 2, if $a$ is any fixed element of $G$, then
$$\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = \{1\}.$$
But here $G$ is abelian, so $\sigma G_a \sigma^{-1} = G_a$ for all $\sigma \in G$, thus
$$G_a = \bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = \{1\}.$$
Let $\sigma \in G - \{1\}$. Since $G_a = \{1\}$, $\sigma 
ot \in G_a$, so $\sigma\cdot a = \sigma(a) 
e a$. This is true for any $a \in A$, so
$$\forall \sigma \in G - \{1\},\ \forall a \in A,\ \sigma(a) 
e a.$$
Since $G$ acts transitively on $A$, there is only one orbit, which is $A$, so the orbit of $a$ is
$${\mathcal O}_a =  A.$$
Since $G_a= \{1\}$, the orbit-stabilizer formula (Proposition 2) gives
$$|A| = |{\mathcal O}_a| = |G : G_a| = |G : \{1\}| = |G|,$$
so
$$|G| = |A|.$$
(In particular, every abelian, transitive subgroup of $S_n$ has $n$ elements.)
\end{proof}

Exercise 4.1.3 (Abelian transitive subgroups of $S_A$)

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Exercise 4.1.3 (Abelian transitive subgroups of $S_A$)

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