Timeline for For a new operation on a finite group of odd order giving a loop structure, when does this also gives a group

Current License: CC BY-SA 3.0

14 events

when toggle format	what		by	license	comment
Jan 16, 2016 at 10:37	history	edited	Geoff Robinson	CC BY-SA 3.0	typo
Jan 16, 2016 at 9:34	history	edited	Geoff Robinson	CC BY-SA 3.0	added final comment
Jan 14, 2016 at 9:02	history	edited	Geoff Robinson	CC BY-SA 3.0	typo, and addressed the issue of nilpotence class
Jan 14, 2016 at 8:48	history	edited	Geoff Robinson	CC BY-SA 3.0	typo
Jan 14, 2016 at 0:49	history	edited	Geoff Robinson	CC BY-SA 3.0	Added proof that $G$ must be nilpotent if $(G,\circ)$ is a group.
Jan 14, 2016 at 0:02	history	edited	Geoff Robinson	CC BY-SA 3.0	Removed remarks on necessary condition in general case- can be simplified.
Jan 13, 2016 at 23:48	history	edited	Geoff Robinson	CC BY-SA 3.0	simplified part of end argument
Jan 13, 2016 at 22:43	history	edited	Geoff Robinson	CC BY-SA 3.0	Added necessary general condition for $(G,\circ)$ should be a group.
Jan 13, 2016 at 10:54	vote	accept	StefanH
Jan 12, 2016 at 6:39	comment	added	M. Farrokhi D. G.		The commutativity is much easier to prove. Indeed, $x\circ y=(x^{\frac{1}{2}}y^{\frac{1}{2}})(y^{\frac{1}{2}}x^{\frac{1}{2}})$ and $y\circ x=(y^{\frac{1}{2}}x^{\frac{1}{2}})(x^{\frac{1}{2}}y^{\frac{1}{2}})$. Since $[x^{\frac{1}{2}}y^{\frac{1}{2}},y^{\frac{1}{2}}x^{\frac{1}{2}}]=[y^{\frac{1}{2}}x^{\frac{1}{2}}[x^{\frac{1}{2}},y^{\frac{1}{2}}],y^{\frac{1}{2}}x^{\frac{1}{2}}]=1$ we have $x\circ y=y\circ x$.
Jan 12, 2016 at 2:18	history	edited	Geoff Robinson	CC BY-SA 3.0	Proved associativity
Jan 12, 2016 at 1:16	history	edited	Geoff Robinson	CC BY-SA 3.0	typo
Jan 12, 2016 at 1:05	history	edited	Geoff Robinson	CC BY-SA 3.0	clarified
Jan 12, 2016 at 1:00	history	answered	Geoff Robinson	CC BY-SA 3.0