Why additional constraint is need for this two groups to be isomorphic? I'm reading AMS's book Papers on Topology, which collects Poincare's papers on topology.
However, the first paper stops me.
In the paper, he considered the group generated by transformations in $\mathbb{R}^3$, the generators are:
$(x,y,z)\rightarrow (x+1,y,z)$
$(x,y,z)\rightarrow (x,y+1,z)$
$(x,y,z)\rightarrow(\alpha x+\beta y,\gamma x+\delta y, z+1)$
where $\alpha\delta-\beta\gamma=1$.
Obviously, the third transformation is a rotation in $XY$-plane and one move in $Z$.
We denote the group generated above by $(\alpha,\beta,\gamma,\delta)$.
Then Poincare claims that two groups $(\alpha,\beta,\gamma,\delta)$ and $(\alpha',\beta',\gamma',\delta')$ cannot be isomorphism unless the two transformation in $\mathbb{R}^2$:
$(x,y)\rightarrow(\alpha x+\beta y,\gamma x+\delta y)$
$(x,y)\rightarrow(\alpha' x+\beta' y,\gamma' x+\delta' y)$
are conjugate of each other by a linear transformation with integer coefficient.
I'm confused by this claim and can not find why.
I think if we denote the three transformation $\sigma_1,\sigma_2,\sigma_3$ and $\sigma_1',\sigma_2',\sigma_3'$, we can just form an isomorphism: $\phi(\sigma_i)=\sigma_i'$.
Why the additional constraint is needed?
 A: Poincare was correct. These are Abelian ($\mathbb{Z}^2$)-by-cyclic groups where the cyclic group acts on $\mathbb{Z}^2$ by the matrix $\left(\begin{array}{ll}\alpha & \gamma \\\ \beta & \delta\end{array}\right)$. These groups are isomorphic if and only if the matrices are conjugate in $SL(2,\mathbb{Z})$. 
Here is more about the proof. Consider two groups $G_1$ and $G_2$ which are extensions of normal subgroups $H_1, H_2$ which are isomorphic to $\mathbb{Z}^2$ by cyclic groups $\langle c_i\rangle$, $i=1,2$ acting (by conjugation) as matrices $M_1, M_2$ from $SL_2(\mathbb{Z})$. Suppose that there exists an isomorphism $\phi$ from $G_1$ onto $G_2$. Assume that the eigenvalues of $M_1$ and $M_2$ are not on the unit circle (otherwise the situation is also easy but somewhat different). Then prove that $\phi$ must take $H_1$ onto $H_2$ because $G_i$ has only one maximal normal Abelian subgroup isomorphic to $\mathbb{Z}^2$. We can assume then that $H_1=H_2$ and $\phi$ acts as a matrix $N$ on $H_1$. Then $N$ conjugates $M_1$ to $M_2$ (from the definition of homomorphism). I do not know how Poincare proved it (in his paper, there is no proof or reference). Perhaps his proof was more geometric. One can also use more modern technique like rigidity theorems. 
A: Since you are interested in topology, you would do well to read Peter Scott's beautiful paper from 1983, called "The geometries of 3-manifolds". The manifolds you are asking about are Solvmanifolds (at least in the case @Mark Sapir is discussing in his answer), and are discussed in detail starting on page 470 of that paper (including an answer to your question), but you would do well to read the whole thing.
