A group which is $\mathbb{Z}$-by-finite but not finite-by-$\mathbb{Z}$ I found a lemma which states: 

If a group $G$ is finite-by-$\mathbb{Z}$, then $G$ is
  $\mathbb{Z}$-by-finite.

I was wondering if the converse is true, i.e. is it true that

if a group G is $\mathbb{Z}$-by-finite, it's also finite-by-$\mathbb{Z}$?

I suspect not, but as I'm only starting to begin to understand the theory, I can't come up with a counterexample. 
Thanks in advance!

EDIT: A brief explanation of $\mathcal{P}$-by-$\mathcal{Q}$:  Let $\mathcal{P}$ and $\mathcal{Q}$ be 2 properties of groups. Then a group $G$ is $\mathcal{P}$-by-$\mathcal{Q}$ if there exist a normal subgroup $N$ of $G$ such that $N$ has property $\mathcal{P}$ and $G/N$ has property $\mathcal{Q}$.
The line "has property $\mathbb{Z}$" means: being isomorphic to the infinite cyclic group.
 A: Let $G$ be the infinite diedral group: $$G=\langle a,b\mid b^2=1, bab=a^{-1}\rangle.$$ The subgroup generated by $a$ is normal and isomorphic to $\mathbb{Z}$, the quotient is of order $2$. Hence, $G$ is $\mathbb{Z}$-by-finite.
However, there is no normal subgroup of $G$ which is finite. So the group is not finite-by-$\mathbb{Z}$. To see that, take a normal subgroup $H\subset G$ and a non-trivial element $h\in H$. If $h\in \langle a\rangle$, then $h$ is of infinite order, so $H$ also. Otherwise, $h=a^lb$ for some $l\in \mathbb{Z}$. You compute then $$aha^{-1}h^{-1}=a^{l+1}ba^{-1}ba^{-l}=a^{l+1}aa^{-l}=a^2\in H$$
and find again that $H$ is of infinite order.
A: The infinite dihedral group $D := \langle x, y : y^{-1}xy = x^{-1}, y^2 = 1 \rangle$ is $\mathbb{Z}$-by finite. 
Suppose $F$ is a finite normal subgroup of $D$. Then $F \cap \langle x \rangle$ is trivial so $D$ is isomorphic to a subgroup of $D / \langle x \rangle \cong C_2$. If $|F| = 2$ then $F = \langle a \rangle$ for some central element $a \in D$ of order two. However the centre of $D$ is trivial, so in fact $|F| = 1$.
Thus $D$ has no non-trivial finite normal subgroups, so its only infinite homomorphic image is $D$ itself. In particular, $D$ cannot be finite-by-$\mathbb{Z}$ since $D$ is not isomorphic to $\mathbb{Z}$ (having a non-trivial element of order two).
A: The dihedral group is in a sense the only exception: a group is $\mathbf{Z}$-by-finite if and only if it is finite-by-$\mathbf{Z}$ or finite-by-$D_\infty$.
