If $G$ has a normal subgroup of order 2 and infinite cyclic quotient, $G$ is abelian?

Assume that $G$ has a normal subgroup $H$ of order $2$ (isomorphic to $Z_{2}$) and $G/H$ is infinite cyclic (which indicates that $G$ is also infinite order). The target here is to prove that $G$ is isomorphic to $Z\times Z_{2}$. And it's pretty obvious that we are going to show $G \cong G/H \times H$ in order to get the conclusion. Under such a context, is $G$ an abelian group?

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1 Answer

Let $H$ be of order $2$; let $g\in G$ be an element that projects onto the generator of $G/H$, and let $K=\langle g\rangle$. Then $K\cap H=\{1\}$ (since the image of $K$ is infinite cyclic); and $KH=G$ by the isomorphism theorems; hence $G$ is a semidirect product $\mathbb{Z}_2\rtimes K$; so $K$ acts by conjugation on $\mathbb{Z}_2$, but that action must be trivial (since $\mathrm{Aut}(\mathbb{Z}_2)$ is trivial); hence $ghg^{-1}=h$ for every $h\in H$, so $G$ is abelian and isomorphic to $H\times K$.

Or somewhat simpler: let $H=\{1,h\}$, and let $g\in G$. Since $H$ is normal, $ghg^{-1}\in H$, and since $ghg^{-1}\neq 1$, then $ghg^{-1}=h$. Thus, $h$ is central, so $H\subseteq Z(G)$. It is well known that if $N\subseteq Z(G)$ and $G/N$ is cyclic, then $G$ is abelian. (Though this does not tell you directly that $G\cong \mathbb{Z}\times\mathbb{Z}_2$.)

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Thanks! I've a hard time visualize why does $KH$=$G$ since I haven't learnt the 1st isomorphism Theorem yet, but I will take a look at it. Thanks again! – juxtex Apr 7 '12 at 21:02
@juxtex: Let $x\in G$. Then $xH\in G/H$, so there exists $k$ such that $xH = g^kH$ (since $\langle gH\rangle=G/H$). Therefore, $x^{-1}g^k=h\in H$, so $x=g^kh^{-1}\in KH$. Thus $G\subseteq KH\subseteq G$ (note that since $H$ is normal, $KH=HK$, so $KH$ is a subgroup). – Arturo Magidin Apr 7 '12 at 21:04