I have read these questions:
Show that $\sigma(a)\ne a,\forall\sigma\in G-\{1\}$ and all $a\in A$.where $G$ is abelian, transitive subgroup of $S_A$,
Show that any abelian transitive subgroup of $S_n$ has order $n$,
but I didn't get what I was looking for. This question is from Dummit & Foote Section 4.1 Exercise 3.
Assume that $G$ is an abelian, transitive subgroup of $S_A$. Show that $\sigma(a)\neq a$ for all $\sigma \in G-\{1\}$ and all $a \in A$. Deduce that $|G|=|A|$.
My proof goes as follows. Take any $a\in A$, and by Proposition 2 of the same section(which states that the number of elements of the orbit is same as the index of the stabilizer in the group), we have $$|G:G_a|=\frac{|G|}{|G_a|}=|O_a|=|G|$$
where $G_a$ is the stabilizer of $a$, $O_a$ is the orbit containing $a$, and the last equality follows from the fact that $G$ acts transitively on $A$. Hence, $|G_a|=1$ and $G_a=\{1\}$. Now take any $\sigma \in G-\{1\}$. Then, $\sigma \not\in G_a$, so we have $\sigma(a)\neq a$. As the choice of both $\sigma$ and $a$ was arbitrary, the first part of the question is solved.
Now, where have I done wrong? I think I'm wrong because I haven't used the abelian property of $G$.
Best Answer
Suppose there is $a\in A$ such that $\sigma(a)=a$ for some $\sigma\in G$. Since $G$ is abelian: $$\forall\tau\in G:\sigma(\tau(a))=\tau(\sigma(a))=\tau(a) \tag1$$ By the transitivity of $G$, as $\tau$ runs in $G$, $\tau(a)$ spans the whole $A$. Therefore, $(1)$ reads: $$\forall b\in A:\sigma(b)=b$$ namely $\sigma=1_G$. Therefore, if $\sigma\ne 1_G$, then $\sigma(a)\ne a$ for every $a\in A$. In turn, this implies that $G_a$ is trivial for every $a\in A$. By the orbit-stabilizer theorem, the only orbit, namely $A$, has then cardinality $|G:G_a|=|G|$.