Timeline for Any method to detect subgroup generated by a subset of the generators from its presentation
Current License: CC BY-SA 3.0
10 events
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| May 21, 2012 at 15:23 | comment | added | ADL | In general, you can use the reidemeister-schreier algorithm to compute the presentation of a subgroup of a given subgroup $H\leq G$. It works by looking at the cosets $G/H$, so it is nicest if $H\lhd G$. If $H$ has finite index in $G$ then it will spit out a finite presentation for $H$. I found a worked example of this here, math.stackexchange.com/questions/59273/… | |
| May 21, 2012 at 15:15 | comment | added | Steve D | @Xiaolei: The answer is certainly yes, as you can see from the presentation I gave. | |
| May 21, 2012 at 14:55 | comment | added | Xiaolei Wu | Agol, Can you tell me why when it is a HNN extension, the answer should be yes? As a matter of fact, my group arise from some topology construction, if I calculated the fundamental group right, it is an HNN extension. | |
| May 21, 2012 at 13:27 | comment | added | Xiaolei Wu | Hi, Mosher, I want to know what the subgroup is or can you find a presentation of this group. | |
| May 21, 2012 at 13:22 | comment | added | Lee Mosher | For your general question, you will have to say more about what type of "detection" you have in mind. If you mean "is there an algorithm to determine membership in the subgroup generated by a subset of the generators", the answer is NO, there is no such algorithm. | |
| May 21, 2012 at 7:08 | comment | added | Steve D | If you call $z=x^y$, then what you've written implies $x^t=z^t=xz$ (because $y^2$ is central in $\langle x,y\rangle$). This impliex $x=z$, so that in fact $y$ is central in $\langle x,y\rangle$, and your group is just a regular HNN extension $\langle x,y,t\ |\ [x,y]=1, x^t=x^2, y^t=y^2 \rangle$. | |
| May 21, 2012 at 6:06 | comment | added | Ian Agol | I meant "isomorphic to H". | |
| May 21, 2012 at 6:05 | comment | added | Ian Agol | Your group might be an HNN extension, in which case the answer to your question should be yes. In the group $H=\langle x,y| xy^2=y^2x\rangle$, you need to determine if the subgroup generated by $\langle y^2, xy^{-1}xy\rangle$ is isomorphic to $G$. Then $G$ is an ascending HNN extension of $H$ with respect to the homomorphism sending $x\mapsto xy^{-1}xy, y\mapsto y^2$. | |
| May 21, 2012 at 5:50 | history | edited | Xiaolei Wu | CC BY-SA 3.0 |
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| May 21, 2012 at 5:44 | history | asked | Xiaolei Wu | CC BY-SA 3.0 |