Given a group $G$, with $|G|=17 \cdot 2 \cdot 3 $ and $Z(G)=2$ show that $|Z(G/Z(G))|=1$ I'm really having trouble finding this one...
Given a group $G$, with $|G|=17 \cdot 2 \cdot 3 $ and $|Z(G)|=2$ show that $|Z(G/Z(G))|=1$ 
Hope you guys can help...
Thanks!
 A: A group having those properties doesn't exist. 
$G/Z(G)$ has $17\cdot3$ elements, meaning it is cyclic, so $|Z(G/Z(G))|=17\cdot3$. In turn, since you modded out by the center and got a cyclic quotient, the whole group must be abelian, a contradiction to your statement that $|Z(G)|=2$. So no such group exists.
A: Listing a few bits. By Cauchy's theorem the group $G$ has elements, call them $x,y,z,$ of respective orders $17$, $3$ and $2$. By Sylow's theorems, group $G$ has a unique subgroup $P$ of orders $17$, the one generated by $x$. As it is unique, we have $P\lhd G$. 
The next thing we observe that $Aut(P)$ is cyclic of order sixteen. Therefore it has no automorphisms of order three. The mapping $f:x^i\mapsto yx^iy^{-1}$ is an automorphism that satisfies $f^3=1_{Aut(P)}$, so $f$ has to be the identity mapping. Therefore $x^i=yx^iy^{-1}$ for all $i$. In particular $x$ and $y$ commute, and together generate a (cyclic) group $H$ of order $51$.
Because $[G:H]=2$, we conclude that $H\lhd G$. Therefore $H$ is stable under conjugation by $z$. Clearly $zxz^{-1}$ is of order $17$, so $zxz^{-1}=x^k$ for some $k$. As $x=z^2xz^{-2}=z(zxz^{-1})z^{-1}=zx^kz^{-1}=x^{k^2}$ we get that $k^2\equiv1\pmod{17}$. So $k\equiv\pm1\mod {17}$. Similarly $zyz^{-1}$ is in $H$, and of order three, so it has to be either $y$ or $y^{-1}$.
Let's take stock. There are four possibilities: $zxz^{-1}=x^{\pm1}$ and
$zyz^{-1}=y^{\pm1}$. If we have a plus sign in both places, then $G$ is abelian, so $Z(G)=G$ and therefore $|Z(G/Z(G))|=1$. If we have a minus sign in both places, then the center of $G$ is trivial, and again $|Z(G/Z(G))|=1$. In this case we have $G\cong D_{51}$.
If the signs differ, then we get something interesting. If $z$ commutes with $x$, but does not commute with $y$, then $z$ is in the center. We see that $G\cong C_{17}\times S_3.$ In this case $Z(G)=P$, and again $Z(G/P)=Z(S_3)=1.$
If $z$ commutes with $y$ but does not commute with $x$, then we get that $G\cong C_3\times D_{17}$. In that case $Z(G)=\langle y\rangle$, and again
$Z(G/Z(G))=Z(D_{17})=1$.
So we have $Z(G/Z(G))=1$ for all groups $G$ of this order.
