Does there exist a simple nonabelian 2generated group $G$ and two elements $a, b \in G$, such that $\langle \{a, b\} \rangle = G$, $a^2 =1$ and $\forall c, d \in G$ $\langle \{c^{1}bc, d^{1}bd \} \rangle \neq G$?

1$\begingroup$ Duplicate of math.stackexchange.com/questions/3600219/… $\endgroup$ – verret Mar 29 at 18:50
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 nonAbelian, $G = \langle b, a^{1}ba \rangle .$

1$\begingroup$ Why $\langle b ,a^{1}ba \rangle$ is normalized by $b$ ? $\endgroup$ – amir bahadory Mar 29 at 18:33

1$\begingroup$ Any subgroup $H$ containing an element $b$ of an overgroup $G$ is certainly normalized by $b$ since $b^{1}Hb = H$ (since $b \in H$). $\endgroup$ – Geoff Robinson Mar 29 at 18:40

$\begingroup$ also we must show why $\langle b, a^{1}ba \rangle \neq \{1\} $ ? I think because $ b \neq 1$. $\endgroup$ – amir bahadory Mar 29 at 21:35

$\begingroup$ Yes, it is of course since $b \neq 1$, but this is clear since $G \neq \langle a \rangle$ (for $G$ is assumed nonAbelian simple). $\endgroup$ – Geoff Robinson Mar 29 at 21:55

$\begingroup$ or If we show for every $n \in \mathbb{N} , b^n \neq 1$ then our subgroup is not identity? $\endgroup$ – amir bahadory Mar 29 at 22:02