# If $|G_1|=|G_2|<\infty$ and $|G_1'|<|G_2'|$, then $|Z(G_1)|\geq |Z(G_2)|$? where $G'$ is the commutator subgroup of $G$.

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 $G/Z(G)\cong \text{inn }G$.

• I can think of counterexamples of order $72$ ($G_1$ with $|G_1'|= 9$, $|Z(G_1)|=1$, $G_2$ with $|G_2'|=18$, $|Z(G_2)|=2$), but there might be smaller examples. – Derek Holt Mar 14 '16 at 9:20
• @DerekHolt What are the groups $G_1$ and $G_2$? Thanks. – bfhaha Mar 14 '16 at 11:05
• There are some clue. mimuw.edu.pl/~zbimar/small_groups.pdf – bfhaha Mar 14 '16 at 11:30
• $G_1 = {\rm AGL}(1,9)$, $G_2 = D_{72}$. – Derek Holt Mar 14 '16 at 11:32
• @DereHolt I got a smaller example. (In fact, it is the smallest such groups.) They are $G_1=\Bbb{Z}_2\times D_8$ and $G_2=(\Bbb{Z}_2\times D_4)\rtimes \Bbb{Z}_2$ or $G_2=(\Bbb{Z}_2\times Q_8)\rtimes \Bbb{Z}_2$. It is the answer from GAP. I am trying to prove that. Thanks. – bfhaha Mar 15 '16 at 5:07