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Oct 5, 2019 at 14:58 comment added YCor Ah sorry I was giving a counterexample to the converse (you wrote "I have tried to do the same", but it's not the same).
Oct 5, 2019 at 14:55 comment added J.L. @YCor but if $h_{1}\in L$, I would obtain that $L$ is generated by $\langle l_{1},\dots,l_{k},h_{1}$, so I have what I want
Oct 5, 2019 at 14:50 comment added YCor Note that in your second paragraph you're stating something for some very particular instance of HNN extension (HNN with respect to identity automorphism of some given subgroup). For the question, the generality is a bit hopeless, as $D$ is likely not be mapped injectively without further assumption. For instance if $h_1\in L$ and $h_1h_2^{-1}$ generates normally $D$ then $D$ is killed in $H$.
Oct 5, 2019 at 14:09 history asked J.L. CC BY-SA 4.0