Infinite Simple Groups

Question

I have been reading about the Rado Graph (or Random graph), which we will call $R$, and had a question regarding its automorphism group, $\operatorname{Aut}(R)$. In a paper that I am reading, it states that $\operatorname{Aut}(R)$ is simple but it omits the proof. Instead the author gives a citation for the following fact:

if any $g,h \in \operatorname{Aut}(R)$ are non-identity elements, then $h$ can be expressed as a product of five conjugates of $g$ or $g^{-1}$,

and then parenthetically says (clearly this implies simplicity). However this is not too clear to me. What little knowledge I have a simple groups is almost all in regards to finite groups. Could someone explain how simplicity follows from the fact above?

Answer 1

Suppose $N\le Aut(R)$ were normal and contained a nonidentity element $g$. Then every conjugate of $g$ and $g^{-1}$ is contained in $N$, since $N$ is normal. But since $N$ is closed under the group operation, any other non-identity $h$ is also contained in $N$. So $N= Aut(R)$.

Answer 2

The statements says any element is a product of 5 conjugates of any non-identity element $g$ or $g^{-1}$. So if $g$ were in a normal subgroup, then all its conjugates can generate any element of $G$. But by normality these have to be in that normal subgroup. So whole of $G$ is in any non trivial normal subgroup. That is, the group is simple.