# Group $G$ whose center $Z(G)$ is cyclic and with $G/Z(G)$ commutative

I have some issue to solve following exercise. The exercise is from a French book on Algebra (cours d'Algèbre) from Jean Querré. The book is from the 1970's.

If the center $Z(G)$ of a group $G$ is cyclic and $G/Z(G)$ is commutative then there exists a group $H$ such that $H \times H$ is isomorphic to $G/Z(G)$.

• Is $G$ finite or possibly infinite? – Matt Samuel Jun 28 '15 at 2:31
• In particular case , If $G$ be abelian then $Z(G)=G$ and $H=\{1\}$.(in this case $Z(G)$ is not nessecery cyclic) – 1200785626 Jun 28 '15 at 2:52
• @Matt Samuel Nothing special is mentionned in the exercise regarding the group cardinal. – mathcounterexamples.net Jun 28 '15 at 13:01
• The paper Groups with preassigned central and central quotient group by Baer should be useful, see corollary 7.2. – Mikko Korhonen Jun 29 '15 at 11:50

This answer is almost complete and I think (read hope) that it's correct up to the point where I get stuck. (I've posted this as an answer because it's too long for a comment and I should go to bed.)

It suffices by the first isomorphism theorem to find a group $H$ and a surjective homomorphism $\theta : G \to H \times H$ for which $\ker \theta = Z(G)$.

By spelling out the definitions it is easy to see that $G/Z(G)$ is commutative if and only if $g^{-1}h^{-1}gh \in Z(G)$ for all $g,h \in G$, which occurs if and only if $[G,G] \le Z(G)$, where $[G,G]$ is the commutator subgroup of $G$.

This implies that the map $G \to [G,G]$ defined by $x \mapsto g^{-1}x^{-1}gx$ is a homomorphism for fixed $g \in G$, since if $y \in G$ then \begin{align} (g^{-1}x^{-1}gx)(g^{-1}y^{-1}gy) &= g^{-1}(x^{-1}gxg^{-1})y^{-1}gy \\ &= g^{-1}y^{-1}(x^{-1}gxg^{-1})gy \\ &= g^{-1}y^{-1}x^{-1}gxy \\ &= g^{-1}(xy)^{-1}g(xy) \end{align} Likewise the map $x \mapsto x^{-1}h^{-1}xh$ is a homomorphism for fixed $h \in G$.

Since $Z(G)$ is cyclic, so is $[G,G]$, and hence it is generated by a single element $g^{-1}h^{-1}gh$ for some $g,h \in G$.

Fix $g,h \in G$ such that $[G,G] = \langle g^{-1}h^{-1}gh \rangle$. Let $H = [G,G]$ and define $\theta : G \to [G,G] \times [G,G]$ by $$\theta(x) = (g^{-1}x^{-1}gx, x^{-1}h^{-1}xh)$$ Now

• We know $\theta$ is a homomorphism by above remarks.
• $\theta$ is surjective. Indeed, if $a \in [G, G] \times [G, G]$ then there exist $m,n$ such that $$a = ( (g^{-1}h^{-1}gh)^m, (g^{-1}h^{-1}gh)^n )$$ But $\theta(h) = ( g^{-1}h^{-1}gh, 1 )$ and $\theta(g) = ( 1, g^{-1}h^{-1}gh )$, so $$a = \theta(h)^m\theta(g)^n = \theta(h^mg^n)$$ hence $a \in \mathrm{im}\, \theta$.
• (This is where I get stuck.) We need to show that $\ker \theta = Z(G)$. The $\supseteq$ direction is obvious. For $\subseteq$, suppose $x \in \ker \theta$ and let $y \in G$. We need to show $xy=yx$. Well $x \in \ker \theta$ implies $xg=gx$ and $xh=hx$, so $x$ commutes with $g$ and $h$ (and their inverses). I'm sure a bit more hacking would yield $xy=yx$ but I haven't quite managed to do it. The fact that $x^{-1}y^{-1}xy \in [G,G]$ and hence $x^{-1}y^{-1}xy = (g^{-1}h^{-1}gh)^n$ for some $n$ should be useful.
• A $p$-group $P$ is called special if its center $Z(P)$, derived subgroup $P'$ and Frattini subgroup $\Phi(P)$ are all equal. If this subgroup is additionally cyclic, then $P$ is called extraspecial. As there are extraspecial groups of order $p^n$ for all odd n and primes $p$, you won't be able to finish your proof with $H = G'$. Try looking at a maximal abelian subgroups $A$ of $G$ instead, and see if $H = A/Z(G)$ works. – j.p. Jun 28 '15 at 14:16
• On second thought, taking a maximal abelian subgroup won't work. Instead assuming $G$ finite, I'd reduce to the case $G$ being a $p$-group. $(g, h)\mapsto [g, h]$ induces a non-degenerate skew-symmetric bilinear map $G/Z\times G/Z \to G'$. I'd induct on the order of $G/Z$ by finding a matching element for an element of $G/Z$ of maximal order. – j.p. Jun 29 '15 at 15:29
• I think the finite case, and even the finitely generated case, is relatively easy. For the general case you probably need to take a maximal subgroup that does decompose in the required way and then show it has to be the whole group. I tried to write it down but I didn't get the details right. – Derek Holt Jun 30 '15 at 7:33
• @DerekHolt: Any chance that you'll give it a second try? The infinite case would interest me. Thanks! – j.p. Jul 10 '15 at 16:24