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Nov
5 |
revised |
Quotients of an extension of the Higman group
added 249 characters in body |
Nov
4 |
revised |
Quotients of an extension of the Higman group
added 134 characters in body |
Nov
3 |
revised |
Quotients of an extension of the Higman group
added 174 characters in body |
Nov
3 |
revised |
Quotients of an extension of the Higman group
added 169 characters in body |
Nov
1 |
revised |
Normal subgroups of an extension of the Higman group
added 206 characters in body |
Oct
31 |
asked | Quotients of an extension of the Higman group |
Oct
30 |
comment |
Normal subgroups of an extension of the Higman group
Another example to work on (given that I seem to have ungraciously broken yours): what about $atatat^2$? I suspect its closure does contain $H$. Is the same true for every word of the form $a^{k_1} t a^{k_2} t a^{k_3} t^2$? |
Oct
30 |
comment |
Normal subgroups of an extension of the Higman group
In the last step, don't you get $H/N = \langle a,b|b a b^{-1} = a^2, a b a^{-1} = b^2\rangle$? And isn't that trivial? |
Oct
29 |
revised |
Normal subgroups of an extension of the Higman group
added 134 characters in body |
Oct
26 |
revised |
Normal subgroups of an extension of the Higman group
added 373 characters in body |
Oct
26 |
comment |
Normal subgroups of an extension of the Higman group
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 |
comment |
Normal subgroups of an extension of the Higman group
Thanks! I guess you did not use the condition $t^4=e$ at all to obtain $a=e$? |
Oct
24 |
comment |
Normal subgroups of an extension of the Higman group
No, I don't think so; conjugate $t^2$ by $a$, say. |
Oct
24 |
revised |
Normal subgroups of an extension of the Higman group
added 165 characters in body |
Oct
24 |
comment |
Normal subgroups of an extension of the Higman group
Thanks! Well, neither $\{e\}$ nor a group containing $H_4$, then. (I suspect your normal closure does coincide with $2\math{Z}/4\mathbb{Z} |
Oct
24 |
revised |
Normal subgroups of an extension of the Higman group
added 96 characters in body |
Oct
24 |
awarded | Nice Question |
Oct
23 |
comment |
Normal subgroups of an extension of the Higman group
Yes, I mean a semi-direct product. I still don't see quite how that (very general) article helps. And yes, I exclude $k_1=k_2=k_3=0$, but the normal closure of that is all of $G$, simply because the Higman-like group with 3 instead of 4 in the definition is trivial. |
Oct
22 |
asked | Normal subgroups of an extension of the Higman group |
Oct
15 |
awarded | Yearling |