Does there exist a simple non-abelian 2-generated group $G$ and two elements $a, b \in G$, such that $\langle \{a, b\} \rangle = G$, $a^2 =1$ and $\forall c, d \in G$ $\langle \{c^{-1}bc, d^{-1}bd \} \rangle \neq G$?
We know that every group $G$ is isomorphic to a subgroup of the symmetric group acting on $G$ by Cayley's theorem .
So, if our example is finite then we can use the fact that if $G$ is a non-abelian finite, simple group of order $>2$ and $G$ is a subgroup of $S_n$, then $G$ must be a subgroup of $A_n$.
However, the group in question is not necessarily finite.
No. Note that $\langle b ,a^{-1}ba \rangle$ is normalized by $b$, and by $a$. Hence $\langle b, a^{-1}ba \rangle$ is normalized by $\langle a,b \rangle = G$. Since $G$ is simple non-Abelian, $G = \langle b, a^{-1}ba \rangle .$also we must show why $⟨b,a^{−1}ba⟩≠{1}$ ? ( because $b≠1$)