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Let $G$ be a group, which is

  1. 2-step nilpotent
  2. torsion-free
  3. generated by three elements (in a minimal presentation)
  4. 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).

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up vote 1 down vote accepted

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.

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Thanks. I guess that means that even in fairly simple cases (nilpotence class 2, 3 generators, torsion-free), these conditions imply very little about the group. I also guess it's a measure of my mathematical ignorance that makes me state overly strong conjectures. Do you have any idea what happens if I add the condition that the centre be generated by one element, i.e. infinite cyclic? – Earthliŋ Apr 18 '13 at 0:27
I found a comment on this question on MO, which claims that one generator can be chosen to be central. However, I do not understand the reasoning by which the commenter deduced this fact. Could there be a simple proof for the case when the number of generators is 3? – Earthliŋ Apr 18 '13 at 3:09
@user1205935, when I gave my first answer, the condition that the centre $Z(G)$ is cyclic was not there! Now I have given a treatment for that case as well. – Andreas Caranti Apr 18 '13 at 12:55

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.

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Thanks, that's a neat counter-example. I would like to add the condition "torsion-free" to $G$. – Earthliŋ Apr 17 '13 at 15:25
Argghhh! Well, that's a huge added condition... – DonAntonio Apr 17 '13 at 15:26
BTW, what do you mean by "2-step t.f. nilp...."? Is that 2-step thing the same as class number? – DonAntonio Apr 17 '13 at 15:27
Yes, really really sorry I didn't remember to put it into the question. By 2-step I mean that $[a,b]$ is in the centre of $G$, i.e. trivial group "in two steps". – Earthliŋ Apr 17 '13 at 15:30
Ok, so if I didn't misunderstand: $$G'=\gamma_2(G)\le Z(G)=\zeta_1(G)\implies \gamma_3(G)=1\implies$$ the nilpotency class of $\,G\,$ is $\,2\,$ and we're back in the case already treated in my answer... – DonAntonio Apr 17 '13 at 15:48

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