# about simple non-abelian 2-generated group [closed]

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 $$\forall c, d \in G$$ $$\langle \{c^{-1}bc, d^{-1}bd \} \rangle \neq G$$?

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, $$G = \langle b, a^{-1}ba \rangle .$$
• Why $\langle b ,a^{-1}ba \rangle$ is normalized by $b$ ? – amir bahadory Mar 29 at 18:33
• Any subgroup $H$ containing an element $b$ of an overgroup $G$ is certainly normalized by $b$ since $b^{-1}Hb = H$ (since $b \in H$). – Geoff Robinson Mar 29 at 18:40
• also we must show why $\langle b, a^{-1}ba \rangle \neq \{1\}$ ? I think because $b \neq 1$. – amir bahadory Mar 29 at 21:35
• Yes, it is of course since $b \neq 1$, but this is clear since $G \neq \langle a \rangle$ (for $G$ is assumed non-Abelian simple). – Geoff Robinson Mar 29 at 21:55
• or If we show for every $n \in \mathbb{N} , b^n \neq 1$ then our subgroup is not identity? – amir bahadory Mar 29 at 22:02