I'm working on the proof of a theorem from A.D. Anderson, about the simplicity of the group of all homeomorphisms of some "sufficiently set-wise homogeneous spaces". Under some conditions on the space $X$, the theorem say that each element of $G=\text{Homeo}(X)$ is the product of 8 conjugates of $h$ and $h^{-1}$ with $h$ an element of $G$.
My actual question is, how do we deduce from this property the algebraic simplicity of the group $G$?
In the case of $X= \mathbb{Q}$ for example, how can we see it?
I thought about this : if we omit the precision of the order of multiplication of the conjugates, the theorem will be like : $ \forall g, \delta \in G$ there are 8 elements $\alpha_{i}$ in $G$ with $ i \in \lbrace{ 1,...,8\rbrace} $ such that : $g= \prod_{i=1}^{8} \alpha^{-1}_{i} \delta \alpha_{i} $ while $ \delta \in \lbrace{h,h^{-1}}\rbrace $
Now, consider $H \mathrel{\unlhd} G$ , and $\delta \in H\setminus{ \lbrace{ id} \rbrace} $, then it follows that $ \alpha^{-1}_{i} \delta \alpha_{i} \in H$ then from the group structure that $g= \prod_{i=1}^{8} \alpha^{-1}_{i} \delta \alpha_{i} \in H$, as $g$ was arbitary, it means that $H=G$ and then $G$ is simple.
is it correct ?