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Timeline for Infinite quotient of Hurwitz Group

Current License: CC BY-SA 3.0

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Aug 22, 2013 at 2:11 comment added Thomas Yes, I'm just working on the simple (ish) cases. I'll stop when it gets too difficult.
Aug 21, 2013 at 14:12 comment added HJRW "I am currently working through all the groups with two generators". Higman, Neumann and Neumann proved that every countable group embeds in a 2-generator group. So your project may take you a while...
Aug 21, 2013 at 11:39 comment added Thomas Wow, is there a shorter representation of it, or is that the shortest? Also, you said something about computers last time, do you have some group theory program that you are using?
Aug 21, 2013 at 7:26 comment added Dima Pasechnik if you know that there is a central element of order 2 then you know that there is a proper infinite quotient group...
Aug 21, 2013 at 7:11 comment added Derek Holt Sorry I'm in a hurry! It's the commutator $[x,y]$, where (for example) $x=b * a * b * a * b^-1 * a * b * a * b * a * b^-1 * a * b * a * b^-1 * a * b^-1 * a * b * a * b^-1 * a * b^-1 * a * b * a * b^-1 * a$, $y=b * a * b^-1 * a * b * a * b^-1 * a * b^-1 * a * b * a * b^-1 * a * b^-1 * a * b * a * b^-1 * a * b * a * b * a * b^-1 * a * b * a$.
Aug 21, 2013 at 7:10 history edited Thomas CC BY-SA 3.0
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Aug 21, 2013 at 7:06 comment added Thomas Ok then, what I need to know is: What is the central element in terms of a and b? Also, is there a way to prove that there aren't any more quotients?
Aug 21, 2013 at 6:59 comment added Derek Holt This was answered in your earlier question. $G$ has a central element of order 2.
Aug 21, 2013 at 4:46 review Close votes
Aug 21, 2013 at 11:35
Aug 21, 2013 at 4:32 comment added Thomas Why the downvote? The question is perfectly reasonable.
Aug 21, 2013 at 4:14 history asked Thomas CC BY-SA 3.0