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Suppose $G$ is a group with a normal subgroup $H \triangleleft G$ such that $\frac{G}{H}$ is cyclic, $H$ is residually finite and $H$ is finitely generated. Show that $G$ is residually finite.

A group $H$ being residually finite means for all $h \in H$ there exists $N \triangleleft H$ such that $[H,N] < \infty$ and $h \notin N$.


So, i've been working on this forever, I think i'm on the right track but could definitely use some help finishing it up and smoothing out some details.

Let $g \in G$. Then, since $\frac{G}{H}$ is cyclic, $g = z^rh$ for some $z \in G$ and $h \in H$. Since $H$ is residually finite, $\exists$ $N \triangleleft H$ such that $h \notin N$ and $[H,N]=n < \infty$. Since $H$ is finitely generated, it has finitely many subgroups of index $n$. Since $H \triangleleft G$, we have that $N^{z^r} \triangleleft H^{z^r} = H$ for all $r \in \mathbb{N}$; thus $N$ has only finitely many conjugates in $G$. Let $I$ be the intersection of all these conjugates of $N$. Then $[H,I] < \infty$ and $I$ char $H$, and so $I \triangleleft G$

edit: STill trying to finish the proof!!

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    $\begingroup$ Related: mathoverflow.net/questions/78341/… $\endgroup$ Oct 10 '20 at 13:14
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    $\begingroup$ In Andreas' answer to the linked question, $Hu^{m\mathbb{Z}}$ is the set $\{hu^{mi}\mid h\in H, i\in\mathbb{Z}\}$. By the conditions set up there, this is a normal subgroup, and has finite index in $G$. Given any element $g\in G$, the subgroup $H$ and the integer $m$ can be picked so that $g\not\in Hu^{m\mathbb{Z}}$, and the result follows. $\endgroup$
    – user1729
    Oct 10 '20 at 13:45
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    $\begingroup$ Cosets are $hu^jHu^{m\mathbb{Z}}$ where $hH$ are cosets of $N/H$ and $0\leq j<m$. $\endgroup$
    – user1729
    Oct 10 '20 at 13:55
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    $\begingroup$ It is easier to see if you write $hu^j$, as two cosets are equal if $(h_1u^{j_1})^{-1}h_2u^{j_2}\in H$, but then $h_1^{-1}h_2\in H$ and $H$ is $u$-invariant. $\endgroup$
    – user1729
    Oct 10 '20 at 15:08
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    $\begingroup$ This is in Andreas' answer: $uHu^{-1}=H=u^{-1}Hu$. $\endgroup$
    – user1729
    Oct 10 '20 at 16:51
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The fact that $H$ is residually finite is not needed. If $G/H$ is generated by $zH$ then $G$ is generated by $z$ and generators of $H$.

The assumption that $H$ is residually finite implies that $G$ is also residually finite. Since that was the real question the proof is this. If the cyclic factor group is finite then $H$ has finite index in $G$ and so $G$ is residually finite (because every subgroup of finite index of $H$ contains a characteristic subgroup of finite index). If the cyclic factor group is infinite then we have an extension of a finitely generated subgroup by a free group which is residually finite (proved by Baumslag): this extension necessarily splits and a semidirect product of two residually finite finitely generated groups is residually finite.

An alternative proof. Every finite index subgroup of $H$ contains a finite index characteristic subgroup. So $G$ is residually an extension of a finite group by a cyclic group. It remains to show that an extension $G$ of a finite group $N$ by a cyclic group is residually finite. But in this group the derived subgroup is inside $N$ so it is finite. Then the center is of finite index. A finitely generated Abelian group is residually finite and we are done.

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  • $\begingroup$ ugh, typed question wrong. meant to say show $G$ is resid finite. $\endgroup$
    – Nobel Cat
    Oct 10 '20 at 19:29
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    $\begingroup$ Both proofs are correct. $\endgroup$
    – Mark
    Oct 11 '20 at 0:12
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    $\begingroup$ Finite groups have finitely many elements. $\endgroup$
    – Mark
    Oct 11 '20 at 2:20
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    $\begingroup$ When we quotient by the characteristic subgroup of finite index of a group with cyclic quotient that subgroup becomes finite. $\endgroup$
    – Mark
    Oct 11 '20 at 2:23
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    $\begingroup$ That is the $N$ in my answer. $\endgroup$
    – Mark
    Oct 11 '20 at 2:55

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