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I'm really lost with this problem and I really need your help: Let $G=\mathbb{Z}^2\rtimes_A\mathbb{Z}$, and let $H\leq G$ with finite index in G. I have to prove that there is a subgroup $U$ of $\mathbb{Z}^2$ and $l\in\mathbb{Z}$ such that $H\simeq U\rtimes_{A^l}\mathbb{Z}$. Where $A=\left( \begin{array}{ccc} 2 & 1 \\ 1 & 1 \\ \end{array} \right)$ . The semiproduct in $G$ is defined: $(x_1,n_1)(x_2,n_2)=(x_1+A^{n_2}x_2,n_1+n_2)$.

Thanks for you hep

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    $\begingroup$ Here is an outline solution. Denoted the normal subgroup ${\mathbb Z}^2$ of $G$ by $N$ and let $U = H \cap N$. Then $H/U \cong HN/N$ which is a subgroup of finite index $l$ in $G/N \cong {\mathbb Z}$. Since $H/U$ is infinite cyclic, $U$ has a complement $\langle g \rangle$ in $H$, where the action of $g$ on $N$ and hence on $U$ is given by the matrix $A^l$. So $H$ is a semidirect product as required. $\endgroup$ – Derek Holt May 20 '16 at 19:19

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