Choosing central generator for nilpotent group generated by 3 elements Let $G$ be a group, which is


*

*2-step nilpotent

*torsion-free

*generated by three elements (in a minimal presentation)

*such that the centre of $G$ is generated by one element (i.e. $C(G)$ is infinite cyclic).


Is it possible to choose generators such that one of the generators is central?
The two answers below give counter-examples, when $G$ is only assumed to satisfy (1) and (3) or (1), (2), and (3).
 A: Take a non abelian group of order $\,|G|=p^3\,\;,\;p\;$  a prime, with exponent $\,p\,$. In this case, we have (with $\,\Phi(G)=$ the Frattini subgroup of $\,G\,$):
$$G'=Z(G)\;,\;\;\Phi(G)=G^pG'\implies Z(G)=G'\le\Phi(G)$$
and this means that every central element is a non-generator of $\,G\,$ , so the answer to your question is no.
A: My previous answer below was given when condition (4) was not there in OP.
With condition (4) added, the answer is yes. The solution I am giving is surely more complicated than it is necessary.
We know $G / Z(G)$ is torsion-free and abelian. Suppose by way of contradiction that $G/Z(G)$ has rank three.
Fix generators $a, b, c$ of $G$ so that $[a, b] \ne 1$, and a generator $d$ of $G'$, so that $G' = \langle d \rangle \le Z(G)$.
Consider the group homomorphism $G/Z(G) \to G'$ given $x \mapsto [x, a]$. If this maps onto $\{ 1 \}$, we are done.
Writing elements of $G$ as product of powers of $a, b, c$, and elements of $G'$ as powers of $d$, we obtain a group homomorphism
$$
\alpha : \mathbf{Z}^{3} \to \mathbf{Z}.
$$
To make things slightly clearer, extend this to a $\mathbf{Q}$-linear map
$$
\alpha' : \mathbf{Q}^{3} \to \mathbf{Q}.
$$
Do the same with $x \mapsto [x, b]$, obtaining
$$
\beta : \mathbf{Z}^{3} \to \mathbf{Z},
\qquad
\beta' : \mathbf{Q}^{3} \to \mathbf{Q}.
$$
Now $\ker(\alpha')$ and $\ker(\beta')$ are subspaces of $\mathbf{Q}^{3}$ of dimension $2$, so they intersect in a one-dimensional subspace $\langle k \rangle$ of $\mathbf{Q}^{3}$ Clearly a multiple of $k$ will be in $\mathbf{Z}^{3}$, and this corresponds to an element
$$
f = a^{s} b^{t} c^{u} \ne 1
$$
such that $[f, a] = [f, b] = 1$. In particular, $f \in Z(\langle a, b, f \rangle)$.
Since we have taken $[a, b] \ne 1$, we have $u \ne 0$. Then $c^{u} \in \langle a, b, f \rangle$, so
$$
1 = [c^{u}, f] = [c, f]^{u},
$$
which implies $[c, f] = 1$ because $G$ is torsion-free, so that $f$ is central. But then the normal subgroup $\langle a, b, Z(G) \rangle$ has finite index at most $u$ in $G$, and this contradicts the fact that $G/Z(G)$ has rank three.

Previous answer
No. A counterexample is the free group in the variety of groups of nilpotence class at most $2$ ($2$-step nilpotent for you).
To be explicit, start with the free abelian group of rank $4$
$$
H = \langle a_{1}, c_{12}, c_{13}, c_{23} \rangle \cong \mathbf{Z}^{4},
$$
then extend it by the automorphism $a_{2}$ of infinite order such that
$$
a_{1}^{a_{2}} = a_{1} c_{12}, \qquad c_{ij}^{a_{1}} = c_{ij},
$$
and then extend $K = \langle a_{1}, a_{2}, c_{12}, c_{13}, c_{23} \rangle$ by 
the automorphism $a_{3}$ of infinite order such that
$$
a_{1}^{a_{3}} = a_{1} c_{13}, \qquad a_{2}^{a_{3}} = a_{2} c_{23}, \qquad c_{ij}^{a_{1}} = c_{ij}.
$$
In the resulting group $G = \langle a_{1}, a_{2}, a_{3} \rangle$ the derived subgroup is free abelian of rank $3$, $G' = \langle c_{12}, c_{13}, c_{23} \rangle  \cong \mathbf{Z}^{3}$. (This is because $[a_{i}, a_{j}] = a_{i}^{-1} a_{j}^{-1} a_{i} {a_{j}} = a_{i}^{-1} a_{i}^{a_{j}}  = c_{ij}$ for $i < j$.)
But if you take any subgroup of $G$ with three generators, one of which is central, its derived subgroup will be cyclic (of rank $1$), generated by the commutator of the two non-central generators.
For a simpler alternative, consider the abelian group of rank $4$
$$
L = \langle a_{2}, a_{3}, c_{21}, c_{31} \rangle \cong \mathbf{Z}^{4}
$$
and extend it by the automorphism $a_{1}$ such that 
$$
a_{2}^{a_{1}} = a_{2} c_{21}, \qquad a_{3}^{a_{1}} = a_{3} c_{31}, \qquad c_{ij}^{a_{1}} = c_{ij}.
$$
The resulting group $M = \langle a_{1}, a_{2}, a_{3} \rangle$ has derived subgroup $M'= \langle c_{21}, c_{31} \rangle$ of rank $2$, so the same argument applies.
