Timeline for 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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Dec 11, 2015 at 1:54	comment	added	YCor		@FaridAliniaeifard we assume by contradiction that $U$ and $V$ (which are both cyclic of order $p$) are conjugate. So, the given generator $e_{23}$ of $U$ is conjugate to some power of the given generator $e_{23}e_{14}$ of $V$.
Dec 11, 2015 at 1:40	comment	added	Farid Aliniaeifard		First line of paragraph 3: Actually $e_{23}$ is conjugate to $g_1 e_{23}e_{14} g_1^{-1} \ldots g_t e_{23}e_{14} g_t^{-1}$ for some $g_1,\ldots, g_t\in G$. How do we have "$e_{23}$ is conjugate to some power $(e_{23}e_{14})^k (1 \leq k \leq p-1)$"?
Dec 10, 2015 at 23:16	history	answered	YCor	CC BY-SA 3.0