>>11123977Yes, almost always. Let's have a look. For simplicity, I will denote by .
If has index n, then acts by left translation on , a set of cardinal .
Therefore, we get a morphism .
We will prove that it is in fact an isomorphism. First, since both sides have order , it is sufficient that we prove injectivity.
Now, the kernel of this action is the intersection . Indeed, an element acts trivially on if and only if, for each , , ie. .
In particular, the kernel of the action is a normal subgroup of index at least .
But the only subgroup of of index is the trivial group (for , it follows from the simplicity of and the rest is checked by inspection).
Hence, is an isomorphism and maps to the stabilizer of the point .
Now, if is a bijection sending the point to , then the induced isomorphism maps the stabilizer of to the stabilizer of in , ie. the image of the standard embedding .
Finally, the composition is an automorphism of mapping to .
