Suppose is an abelian and transitive subgroup. Show that for all in and all . Deduce that .
For each , we have since is abelian. So for all and , we have .
Because the action of is transitive, if we fix then for every there exists such that . Suppose now that ; then , so that . Hence . Thus we can say, for a fixed , that for every there exists a unique such that . In this way we define an injective mapping . Hence .
Now with fixed, we define by . This mapping is injective since if , we have . Thus .