We know that $G'$ characterization how ``abelian'' of a group because we have a theorem: if $G'=\{e\}$, then $G$ is abelian.
I have a conjecture.
If there are two finite groups $G_1$ and $G_2$, $|G_1|=|G_2|$ and $|G_1'|< |G_2'|$, where $G'$ is the commutator subgroup of $G$, does it imply that $|Z(G_1)|\geq |Z(G_2)|$?
Or find a counterexample, is there an example satisfing $|G_1|=|G_2|<\infty$, $|G_1'|< |G_2'|$ and $|Z(G_1)|< |Z(G_2)|$?
This conjecture is true for $|G|\leq 30$ by verify manually myself and this website.
Remarks There is a similar theorem: if $\text{inn }G=\{e\}$, then $G$ is abelian, where $\text{inn }G$ is the group of inner automorphism on $G$.
If there are two finite groups $G_1$ and $G_2$, $|G_1|=|G_2|$ and $|\text{inn }G_1|< |\text{inn }G_2|$, which imply that $|Z(G_1)|> |Z(G_2)|$ because $G/Z(G)\cong \text{inn }G$.