counter example for semi direct product of groups Hi,
I have got a very natural question in group theory.
Suppose you have two countable groups $G_1,G_2$, some action of $\mathbb Z$ on them such that 
the semi direct products are isomorphic  $\phi:G_1\rtimes \mathbb Z\simeq G_2\rtimes \mathbb Z$.
We suppose that $\phi(\mathbb Z)=\mathbb Z$.
Do we have that $G_1\simeq G_2$?
It looks silly but I have not been able to find a counterexample.
Arnaud
 A: Take a compact 3-manifold $M$ with $b_1(M)\geq 2$. Then there are many homomorphisms $\pi_1(M)\to \mathbb{Z}$, since $\mathbb{Z}^{b_1(M)}\leq H_1(M)$. Further, if the manifold fibers over $S^1$ corresponding to a map $\phi:M\to \mathbb{Z}$, then $ker(\phi)$ is finitely generated. If $\phi:M\to \mathbb{Z}$ is not fibered, then a theorem of Stallings implies that the cohomology class is not dual to a fiber. For example, consider the link L4a1:

The complement is a compact manifold $M$ with $H_1(M)=\mathbb{Z}^2$. Orienting the two circles of the link in two different ways (up to negation) gives two different homomorphisms to $\mathbb{Z}$ (via  the linking number). One orientation corresponds to a fibering, while the other does not (there is an annulus running between the two components). Also, the intersection number with the meridian is the same (up to sign) for each choice of orientation, so the cyclic subgroup condition is satisfied. So the kernel of one map is finitely generated (in fact free), while the other is infinitely generated. 
A: No.
Let $X$, $Y$, and $Z$ be infinite cyclic groups with generators $x,y,z$. Make a semidirect product $XY$ using the nontrivial action of $X$ on $Y$. Make the direct product of this with $Z$. In this group there is the free abelian group $XZ$, and inside that there is the infinite cyclic group generated by $xz$. This has two "normal complements" $XY$ and $YZ$. One is nonabelian and the other is abelian.
A: If instead of $\phi(\mathbb{Z}) = \mathbb{Z}$, you know that $\phi$ commutes with the projections to $\mathbb{Z}$, then $\phi$ induces an isomorphism between the kernels of the maps to $\mathbb{Z}$, i.e. between $G_1$ and $G_2$.
But the semidirect products fit into split extensions of $\mathbb{Z}$ by $G_1$ and $G_2$, so it does make sense to ask that $\mathbb{Z}$ is a subgroup of the semidirect product. However I can't quite see how asking $\phi$ to preserve $\mathbb{Z}$ implies that it commutes with the quotient maps.
A: I think this works, and is based on the interpretation of  Tom Goodwillie above.
Let $G = \langle a,b,c\ |\ [a,b]=[a,c]=b^{11}=1,b^c=b^4\rangle$. Now let $H=\langle a^{-1}c^{-1}\rangle$, $G_1=\langle b,c\rangle$ and $G_2=\langle b,ac^2\rangle$.  Then $G=HG_1=HG_2$, but $G_1$ is not isomorphic to $G_2$.
