Let $G$ be a group with center $Z(G)$, and suppose that $N ≤ Z(G)$. Prove by example that $Z(G)/N$ need not be equal to $Z(G/N)$. 
Let $G$ be a group with center $Z(G)$, and suppose that $N ≤ Z(G)$. Prove by example that $Z(G)/N$ need not be equal to $Z(G/N)$.

I'm having trouble coming up with an example. I've tried groups like $S_n$, $D_4$, and $Q_8$ but every time I get that $Z(G)/N = Z(G/N)$. Thanks.
 A: Let $G$ be $Q_8\times Q_8$ where $Q_8=\{\pm 1, \pm i, \pm j, \pm k\}$
is the quaternion group. It has center $Z(G)=\{(\pm1, \pm1)\}$ of order $4$ which has a subgroup $N=\{(\pm1,1)\}$ of order $2$. Then $G/N$ has order $32$ and is a direct product of $Q_8$ and the Klein 4-group, $Z(G/N)$ has order  exactly $8$ but $Z(G)/N$ has order $2$.
A: I don’t know what you did with $Q_8$ and with $D_4$, but in either case, unless you choose $N=\{e\}$, you get an example.
In $Q_8$, if $N=Z(G)=\{1,-1\}$, then $Z(G)/N$ is trivial; but $G/N$ is a group of order $4$, hence abelian, so $Z(G/N)=G/N\neq Z(G)/N$. Same thing in $D_4 = \{r,s\mid r^4=s^2=1,\ sr=r^3s\}$. The center is $\{1,r^2\}$, and if $N=Z(G)$, then $Z(G)/N$ is trivial, but $G/N$ is of order $4$, hence abelian, so $Z(G/N) = G/N \neq Z(G)/N$.
Even if you require $N\neq Z(G)$, these two examples easily give you one satisfying this condition: e.g., take $G=D_4\times C_2$. The center is $Z(G)= Z(D_4)\times C_2$, and taking $N=Z(D_4)\times\{1\}$ works: $G/N$ is abelian, so $Z(G/N)=G/N$, but $Z(G)/N\cong C_2\neq G/N$.
A: A slightly stronger condition on $N$ makes it true: if $N \cap G'=1$, then $N \subseteq Z(G)$ and $Z(G/N)=Z(G)/N$. 
To see this: if $n \in N$ and $g \in G$, then $[g,n]=(g^{-1}n^{-1}g)n \in N \cap G'$, so $[g,n]=1$ for all $n \in N$ and $g \in G$, that is $N \subseteq Z(G)$. If $Z(G/N)=L/N$, then $[L,G] \subseteq N \cap G'=1$. Hence $L \subseteq Z(G)$ and $Z(G/N) \subseteq Z(G)/N$. The reverse inclusion is trivial.
