Timeline for Finite groups which have elements $g$ of order $pq$ such that the sizes of the conjugacy classes of $g^p$, of $g^q$ and of $g^{q-p}$ coincide [closed]

Current License: CC BY-SA 3.0

16 events

when toggle format	what		by	license	comment
Jun 17, 2015 at 19:53	history	closed	Ricardo Andrade Derek Holt András Bátkai Chris Godsil Alex Degtyarev		Not suitable for this site
Jun 17, 2015 at 17:25	history	edited	Stefan Kohl♦	CC BY-SA 3.0	Tried to improve the formulation of the question.
Jun 17, 2015 at 16:53	comment	added	Derek Holt		This was crossposted to math.stackexchange.com/questions/1328886
Jun 17, 2015 at 16:03	answer	added	Geoff Robinson		timeline score: 2
Jun 17, 2015 at 15:49	comment	added	Geoff Robinson		Even though I did misunderstand the question, j.p is correct: since $q-p$ is coprime to $p$ and $q$, it is coprime to $o(g)$, so that $g^{q-p}$ generates $\langle g \rangle$.
Jun 17, 2015 at 15:45	comment	added	Anna		No. I know that $p\neq2$, so $\langle g\rangle\ne\langle g^{q-p}\rangle$
Jun 17, 2015 at 15:40	comment	added	j.p.		@milad: Don't you have $\langle g\rangle = \langle g^{q-p}\rangle$, so both have the same centralizer and therefore their conjugacy classes have the same size?
Jun 17, 2015 at 15:26	comment	added	Stefan Kohl♦		@GeoffRobinson: Is it possible that you misunderstood the question?
Jun 17, 2015 at 15:19	comment	added	Anna		Professor Robinson, you want to say that this equality does not happens, when $G\neq\langle g\rangle$?
Jun 17, 2015 at 15:14	answer	added	Stefan Kohl♦		timeline score: 3
Jun 17, 2015 at 14:57	review	Close votes
Jun 17, 2015 at 19:53
Jun 17, 2015 at 14:57	comment	added	Anna		Why do you consider $G$ is of order $pq$?
Jun 17, 2015 at 14:42	comment	added	Anna		I cannot get your mean. Could you please explain more?
Jun 17, 2015 at 14:39	comment	added	j.p.		I guess you could simplify $(g^{q-p})^G$ to $g^G$.
Jun 17, 2015 at 14:32	review	First posts
Jun 17, 2015 at 14:43
Jun 17, 2015 at 14:32	history	asked	Anna	CC BY-SA 3.0

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