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 $\forall c, d \in G$ $\langle \{c^{-1}bc, d^{-1}bd \} \rangle \neq G$?
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1$\begingroup$ Duplicate of math.stackexchange.com/questions/3600219/… $\endgroup$– verretCommented Mar 29, 2020 at 18:50
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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, $G = \langle b, a^{-1}ba \rangle .$
answered Mar 29, 2020 at 18:27
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1$\begingroup$ Why $\langle b ,a^{-1}ba \rangle$ is normalized by $b$ ? $\endgroup$ Commented Mar 29, 2020 at 18:33
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1$\begingroup$ Any subgroup $H$ containing an element $b$ of an overgroup $G$ is certainly normalized by $b$ since $b^{-1}Hb = H$ (since $b \in H$). $\endgroup$ Commented Mar 29, 2020 at 18:40
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$\begingroup$ also we must show why $\langle b, a^{-1}ba \rangle \neq \{1\} $ ? I think because $ b \neq 1$. $\endgroup$ Commented Mar 29, 2020 at 21:35
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$\begingroup$ 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). $\endgroup$ Commented Mar 29, 2020 at 21:55
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$\begingroup$ or If we show for every $n \in \mathbb{N} , b^n \neq 1$ then our subgroup is not identity? $\endgroup$ Commented Mar 29, 2020 at 22:02