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I'm looking for an example of a non-abelian group $G$ such that $G/Z(G)$ is abelian, where $Z(G)$ is the center of the $G$. In other words, I'm looking for a non-abelian group where $Z(G)$ contains the commutator subgroup $[G,G]$.

Further, what if I request $G/Z(G)$ to be a finitely generated and abelian group?

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  • $\begingroup$ Do you require $G$ to be not finitely generated? If not, the quaternion group $Q=\lbrace\pm1,\pm i,\pm j,\pm k\rbrace$ fits that bill. The center is $\lbrace \pm1\rbrace$, and the quotient by the center has order $4=2\times 2$ so has to be commutative, being the square of a prime number. It's actually isomorphic to $\Bbb Z/2\Bbb Z\times\Bbb Z/2$. $\endgroup$ May 4, 2014 at 12:44
  • $\begingroup$ @OlivierBégassat That's a good example, please post as an answer. $\endgroup$
    – mez
    May 4, 2014 at 12:45
  • $\begingroup$ That is also the smallest example along with the dihedral gro up. $\endgroup$
    – Pedro
    May 4, 2014 at 12:47
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    $\begingroup$ @PedroTamaroff I don't think Dihedral group has a center. $\endgroup$
    – mez
    May 4, 2014 at 12:54
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    $\begingroup$ Well you are mistaken: the centre of the dihedral group of order $8$ has order $2$. $\endgroup$
    – Derek Holt
    May 4, 2014 at 13:29

3 Answers 3

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The quaternion group $Q=\lbrace\pm1,\pm i,\pm j,\pm k\rbrace$ fits that bill. Its center is $\lbrace \pm1\rbrace$, and the quotient by the center has order $4=2\times 2$, so has to be commutative, being the square of a prime number. It's actually isomorphic to $\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$.

EDIT 1. As noted by @DonAntonio, this example generalizes to all finite groups of order $p^3$ (where $p$ is a prime number). Indeed, $p$ groups have non trivial center, and the quotient of a non commutative group by its center can't be cyclic, so that a non commutative group of order $p^3$ must have center of order $p$, and the quotient by its center must be non cyclic of order $p^2$, thus isomorphic to $\Bbb Z/p\Bbb Z\times\Bbb Z/p\Bbb Z$.

EDIT 2. An example of a non commutative group of order $p^3$ is given by the group of uppertriangular matrices of size $3\times 3$ with coefficients in $\Bbb Z/p\Bbb Z$ and a diagonal of ones: $$\left\lbrace\begin{pmatrix}1&a&b\\&1&c\\&&1\end{pmatrix}\left|\right.\, a,b,c\in\Bbb Z/p\Bbb Z\right\rbrace$$ (The group operation is of course given by matrix multiplication.)

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    $\begingroup$ In general, any non-abelian group of order $\;p^3\;,\;\;p$ a prime. In fact, in all these cases we have that $\;G/Z(G)\cong C_p\times C_p\;$ +1 $\endgroup$
    – DonAntonio
    May 4, 2014 at 12:47
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    $\begingroup$ @DonAntonio thanks, I incorporated your remark into my answer. $\endgroup$ May 4, 2014 at 12:56
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A good example of an infinite group of this kind is the "integer Heisenberg group". It has a matrix representation as the group of $3 \times 3$ matrices with $1$'s on the diagonal, $0$'s below the diagonal, and integers above the diagonal. Its center is the infinite cyclic group generated by matrix with one $1$ in the upper right hand corner and with two $0$'s on the super diagonal (and $1$'s on the diagonal). The quotient group is isomorphic to $\mathbb{Z}^2$, represented by nonzero entries on the superdiagonal.

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Consider $D_4 = \{r_0, r_1, r_2, r_3, h, v, d, t\}$. The center is $Z(D_4) = \{r_0, r_2\}$, and the cosets are

$$D_4/Z(D_4) = \Big\{ \;Z(D_4)r_0\;,\; Z(D_4)r_1\;,\; Z(D_4)d\;,\; Z(D_4)h\; \Big \}.$$

Check for yourself through the operation table that all cosets commute, so indeed $D_4/Z(D_4)$ is abelian.

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