An example for a group so that $ Z(G) ≨ G'$

Question

I am trying to find examples for groups so that their commutator subgroup $G'=[G,G]$ has the following relation to the center $Z(G)$:

1. $Z(G) = G'$
2. $Z(G) ≨ G'$
3. $G' ≨ Z(G)$
4. $G' \nsubseteq Z(G)$ and $Z(G) \nsubseteq G'$

(Where $≨$ denotes "a subgroup which is not equal to".) I thought about the Heisenberg group of order $3\times3$ for no. 1, and no. 3 is true for every (non-trivial) Abelian group, as $G' = \{e\}$ while $Z(G)=G$. However, I cannot find a good example for no. 4 or no. 2. Any advice?

Answer 1

1. $G=Q_8=\{\pm 1,\pm i,\pm j,\pm k\}$.Then $G'=Z(G)=\{\pm 1\}$.

2. $G=S_3$. Then $G'=A_3$ and $Z(G)=1$.

3. $G=C_8$. Then $Z(G)=G$ and $G'=1$.

4. $G=S_3\times C_4$. Then $Z(G)\cong C_4$.