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Jun 7, 2016 at 8:01 comment added Geoff Robinson Some of the above relations are redundant: it would be better to write $G = \langle s,t : s^{4} = 1, s^{2} = (st)^{3} \rangle$.
May 8, 2016 at 16:19 comment added user81684 This is just the sort of thing I was looking for; I would have accepted both answers if I could :)
May 8, 2016 at 11:26 history edited Geoff Robinson CC BY-SA 3.0
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May 8, 2016 at 11:15 history edited Geoff Robinson CC BY-SA 3.0
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May 8, 2016 at 11:06 history answered Geoff Robinson CC BY-SA 3.0