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Let $K$ be a group, $G \unlhd K$ be a finite normal subgroup of even order, and let $\langle h \rangle<K$ be an infinite cyclic subgroup, so that they fit into a short exact sequence $$0\to G\to K \to \langle h \rangle \to 0.$$

Question: when can $K$ modulo the normal subgroup generated by $\langle h \rangle$ be trivial?

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  • $\begingroup$ Does "$K/(h)$ trivial" implies K is infinite cyclic? If it's true, how could $G<K$ be normal? $\endgroup$
    – LeechLattice
    Commented Nov 8, 2021 at 8:37
  • $\begingroup$ (h) is a cyclic subgroup of K. I asked when K/H, where H is the normal subgroup generated by (h), is trivial. $\endgroup$
    – piper1967
    Commented Nov 8, 2021 at 8:43
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    $\begingroup$ Yes. There are easy examples. (1) Take $G$ simple non-abelian and any non-trivial action for the semidirect product. (2) Take $G$ abelian of odd order and $h$ acting on $G$ by $x\mapsto -x$. Etc. In general let $u$ be the automorphism of $G$ induced by $h$. Then the required property holds iff the normal subgroup of $G$ generated by $\{g^{-1}u(g):g\in G\}$ is $G$ itself. $\endgroup$
    – YCor
    Commented Nov 8, 2021 at 9:58
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    $\begingroup$ The problem I always have with this type of question is that I have no idea what sort of answer the poster is looking for. All you can really say is that sometimes it is and sometimes it is not. $\endgroup$
    – Derek Holt
    Commented Nov 8, 2021 at 10:14
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    $\begingroup$ @piper1967 I don't have a reference, but this seems to be a straightforward verification. In general, if $N$ is this normal subgroup, the quotient $K/\langle\!\langle h\rangle\!\rangle$ is naturally isomorphic to the finite group $G/N$. $\endgroup$
    – YCor
    Commented Nov 8, 2021 at 10:56

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