For which groups When does $\langle grg^{-1}|g\in G\rangle = \langle gsg^{-1}|g\in G\rangle$ imply $\langle r\rangle=\langle xsx^{-1}\rangle$?

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edited Dec 10, 2015 at 21:45

Arturo Magidin

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For which groups does <grg^$\langle grg^{-1}|g in G>|g\in G\rangle = <gsg^\langle gsg^{-1}|g in G>|g\in G\rangle$ imply <r> = <xsx^$\langle r\rangle=\langle xsx^{-1}> for some x in G\rangle$?

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edited Dec 10, 2015 at 20:53

Farid Aliniaeifard

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We call a finite group $G$ normal if for all $s,r\in G$, if $\langle gsg^{-1}:g\in G\rangle =\langle grg^{-1}:g\in G\rangle $ then there exists $x\in G$ such that $\langle r\rangle =\langle xsx^{-1}\rangle $.

Abelian groups are normal groups and simple symmetric groups $S_n (n\geq 5)$ are not.

Is a $p$-group necessarily normal?

What is the largest family of normal groups?

rephase, clarify.

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edited Dec 10, 2015 at 20:28

Arturo Magidin

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asked Dec 10, 2015 at 18:59

Farid Aliniaeifard

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For which groups When does $\langle grg^{-1}|g\in G\rangle = \langle gsg^{-1}|g\in G\rangle$ imply $\langle r\rangle=\langle xsx^{-1}\rangle$?

For which groups does <grg^$\langle grg^{-1}|g in G>|g\in G\rangle = <gsg^\langle gsg^{-1}|g in G>|g\in G\rangle$ imply <r> = <xsx^$\langle r\rangle=\langle xsx^{-1}> for some x in G\rangle$?