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I have met the following problem. A group $G$ is given as follows

$G = \langle x,y,t| y^{-2}xy^2 = x,t^{-1}yt =y^2 ,t^{-1}xt = xy^{-1}xy\rangle$

Is the subgroup generated by $y$ and $t$ just the group $\langle y,t|t^{-1}yt =y^2 \rangle$? It seems not right, because the first relation and the third relation may be able to give more relations on $y$ and $t$. But I could not prove it.

And the general question would be given a group by a presentation, is there any method to determine the subgroup generated by a subset of the generators?

Sorry, I try to edit the question, but deleted it, so I have to post it again.

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    $\begingroup$ 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 $x\mapsto xy^{-1}xy, y\mapsto y^2$. $\endgroup$
    – Ian Agol
    Commented May 21, 2012 at 6:05
  • $\begingroup$ I meant "isomorphic to H". $\endgroup$
    – Ian Agol
    Commented May 21, 2012 at 6:06
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    $\begingroup$ 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 $\langle x,y,t\ |\ [x,y]=1, x^t=x^2, y^t=y^2 \rangle$. $\endgroup$
    – Steve D
    Commented May 21, 2012 at 7:08
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    $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Commented May 21, 2012 at 13:22
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    $\begingroup$ 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/… $\endgroup$
    – ADL
    Commented May 21, 2012 at 15:23

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