Timeline for Normal subgroups of an extension of the Higman group
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 5, 2015 at 19:51 | history | bounty ended | H A Helfgott | ||
| Oct 26, 2015 at 11:42 | history | edited | Frieder Ladisch | CC BY-SA 3.0 |
typo in formula
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| Oct 26, 2015 at 11:38 | comment | added | Frieder Ladisch | @HAHelfgott: I think the proof of $a=1$ here is still good when we don't have $t^4=1$. In line 5 of this answer, we have $t^{-3} = a^kb^lc^m = \dots$ instead of $t= \dots$, and by the other relations, we also have $b = t^{-3} a t^3$, which can be used instead of $b=tat^{-1}$. The rest of the proof does not depend on $t^4 = 1$, as far as I can see. But we have only the relation $t^3=1$, if we omit $t^4=1$, so we get a cyclic group of order $3$, not the trivial group. | |
| Oct 26, 2015 at 11:10 | comment | added | H A Helfgott | Ah, no, I see you do it tacitly at the very beginning. This makes me curious. If we remove the condition $t^4=e$ (but keep the other relations), must the group still be trivial? | |
| Oct 26, 2015 at 10:43 | comment | added | H A Helfgott | Thanks! I guess you did not use the condition $t^4=e$ at all to obtain $a=e$? | |
| Oct 26, 2015 at 6:19 | history | answered | andrew | CC BY-SA 3.0 |