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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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