# $G$ group, $H \triangleleft G$ such that $\frac{G}{H}$ is cyclic, $H$ is residually finite and $H$ is finitely generated

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!!

• Oct 10 '20 at 13:14
• 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. Oct 10 '20 at 13:45
• Cosets are $hu^jHu^{m\mathbb{Z}}$ where $hH$ are cosets of $N/H$ and $0\leq j<m$. Oct 10 '20 at 13:55
• 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. Oct 10 '20 at 15:08
• This is in Andreas' answer: $uHu^{-1}=H=u^{-1}Hu$. Oct 10 '20 at 16:51

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.

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