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14h
awarded  Critic
1d
revised Free groups of rotations of the sphere
added 49 characters in body
1d
asked Free groups of rotations of the sphere
2d
answered Looking for explanation of Banach-Tarski Proof, preferably by visual methods “Video, Pictures, Diagrams…”
2d
comment free groups of rotations
So that assertion is too strong. What about this: If A and B are noncommuting rotations of infinite order, then G, the group they generate, has a free subgroup of rank 2.
May
19
accepted free groups of rotations
May
19
comment free groups of rotations
Many thanks. It might be nice to have a specific examples of the 3x3 matrices that do not commute but satisfy a word such as you suggested, but I don't really need that. It is enough to know that the theorem is false, and so this adds to the value of the many theorems that say that certain pairs of rotations are independent.
May
18
comment free groups of rotations
30 years on, I am working (with G. Tomkowicz) on revising the BTP book. Big job. Almost finished. That is where this question arose.
May
18
asked free groups of rotations
May
6
answered Solving quadratic or higher degree congruence with very large modulus.
Mar
14
revised Prove that every outer planar graph is $3$-choosable
added 408 characters in body
Mar
14
answered Prove that every outer planar graph is $3$-choosable
Mar
12
answered Examples of non-hamiltonian decomposable graphs
Mar
8
revised Worst case in decanting puzzles (pouring water from one jug to others).
I forgot that I had made the assumption that A0+B0+C0 is at most A, where A0, etc is the initial state.
Mar
7
awarded  Curious
Mar
6
asked Worst case in decanting puzzles (pouring water from one jug to others).
Aug
11
awarded  Yearling
Jun
8
awarded  Good Answer
Dec
21
comment If $X$ is $G$-paradoxical then $G$ is $G$-paradoxical. Is my proof correct?
Sorry if I did not read it in detail, but I guess all is well. And I guess my old hint is acceptable. But, as noted, we are now working on a second edition so if you see anything that is obscure, or could benefit from an elucidating phrase, (or is wrong!) please tell me. You can find my e-address atop by web page, stanwagon.com
Dec
21
answered If $X$ is $G$-paradoxical then $G$ is $G$-paradoxical. Is my proof correct?