Show that $\sigma(a)\ne a,\forall\sigma\in G-\{1\}$ and all $a\in A$.where $G$ is abelian, transitive subgroup of $S_A$

group-actionsgroup-theory

Assume that $G$ is an abelian transitive subgroup of $S_A$. Show that $\sigma(a)\ne a, \forall\sigma\in G – \{1\}$ and all $a\in A$.

I know what transitive subgroup is. I just can't figure out whether the abelian subgroup of $S_A$ will be unique or not.

Best Answer

First thing, $S_A$ certainly doesn't have a unique transitive abelian subgroup in general (even up to conjugation).

Any abelian group of order $|A|$ is in fact isomorphic to a transitive abelian subgroup of $S_A$.

For example $C_2\times C_2\cong\langle (1,2)(3,4),(1,3)(2,4)\rangle$ and $C_4\cong\langle (1,2,3,4)\rangle$.

Now suppose $\sigma(a)=a$. Since $G$ is transitive, for any $b\in S$ there is some $\rho\in G$ with $b=\rho(a)$.

What is $\rho\sigma\rho^{-1}(b)$?

Since $G$ is abelian and the above is true for all $b\in A$ what does this tell you about $\sigma$?