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Mar 7, 2019 at 15:20 vote accept Joakim Færgeman
Mar 6, 2019 at 12:09 answer added Alex B. timeline score: 4
Mar 6, 2019 at 7:59 comment added Derek Holt If $|G|$ is even, then the third condition implies that $p=2$.
Mar 6, 2019 at 2:55 comment added LSpice So $m = 1$ would be all right? (I don't have any examples; I just want to make sure I understand what you're asking.)
Mar 5, 2019 at 22:23 comment added Carl-Fredrik Nyberg Brodda With the risk of stating the obvious: have you tried GAP?
Mar 5, 2019 at 22:22 comment added Joakim Færgeman Thank you for the comments. I have edited the question to answer your comments.
Mar 5, 2019 at 22:21 history edited Joakim Færgeman CC BY-SA 4.0
added 46 characters in body
Mar 5, 2019 at 22:12 comment added Derek Holt I am not finding the question clear. How can you have an even counterexample to a statement about groups of odd order? So you are looking for a group of even order that satisfies the three conditions in the bullet points, is that right? Also you have not said what $m$ is.
Mar 5, 2019 at 22:11 comment added LSpice When you say "an even counterexample", you mean a finite group $G$ of even order for which the three bulleted items do hold (for some $m$ and $p$)? Is it required that $m \ne p$?
Mar 5, 2019 at 21:45 history edited Joakim Færgeman CC BY-SA 4.0
edited title
Mar 5, 2019 at 20:25 history asked Joakim Færgeman CC BY-SA 4.0