2

I want to classify all groups $G$ containing an infinite cyclic subgroup $H$ of index 2. My question is the same as this one, but I was given a hint that the answers in this post didn't completely address.

The hint said to first consider the abelian case. Since $H$ is finitely generated by one element and the subgroup has finite index, it follows that $G$ is finitely generated. So we can use the structure theorem on f.g. abelian groups. I believe it is straightforward from here to deduce the group is isomorphic to either $\mathbb{Z}$ or $\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.

For the non-abelian case, the hint says to consider a natural group action of $G$ on $H$ and use it to show that $G$ contains an element of order $2$. I'm not sure how to proceed here. Since $H$ has index $2$, then $H$ is normal in $G$, so I suppose the natural action of $G$ on $H$ is conjugation by elements of $G$. I don't know how to use this to show the existence of an order $2$ element. Though after this, I believe it's straightforward to conclude that $G$ is isomorphic to $\mathbb{Z} \rtimes \mathbb{Z}/2\mathbb{Z}$ because this order $2$ element gives us a right splitting of the sequence $0 \rightarrow H \rightarrow G \rightarrow \mathbb{Z}/2\mathbb{Z}\rightarrow 0$.

I would appreciate any feedback on how I handled the hints and advice on obtaining this order $2$ element.

Dalop
  • 715
  • 1
    Does it help to observe that $\forall g \in G, g^2 = h^n$ for some $n \in \Bbb Z$, where $H = \langle h \rangle$? – Robert Shore Jun 21 '23 at 18:47

1 Answers1

2

Considering your last case, with $H=\langle h\rangle$ and $g\notin H$. We know that $g^2\in H$, so $g^2=h^n$ for some integer $n$. Also, because the group is nonabelian but $G=\langle H,g\rangle$, it follows that $g$ must act like the unique nontrivial automorphism of $H$, so $ghg^{-1}=h^{-1}$.

Note that if $g\notin H$, then $g^3\notin H$ (it is a product of $g^2\in H$ with $g\notin H$). But $$g^3 = gg^2 = gh^n = gh^ng^{-1}g = h^{-n}g.$$ Cancelling $g$ we get $g^2 = h^{-n}$. Since we also know that $g^2=h^n$, we have $h^{-n}=h^n$, so $n=0$. Thus, $g^2=e$ and hence $g$ is of order $2$, as desired.

Arturo Magidin
  • 417,286