Let $G$ be a group and $H$ be a subgroup of $G$. Classify all pairs $(G,H)$ such that $Z(H)=Z(G)\cap H$ is true.

My attempt:

If $G$ is abelian then $H \le G$ will also be abelian. Hence $Z(H)=H$ and $Z(G)=G$. So $Z(H)=H=G\cap H=Z(G)\cap H$. Also for any group $G$, $H=\{1\}$will also hold trivially. If $H=G$, then $Z(G)=Z(G)\cap G$ is also always true for any $G$, since $Z(G) \le G$

But this is not true for all pairs, consider the quaternion group $G = Q_8$, then $Z(Q_8) = \{1,-1\} $ , and $H = <i> = \{1,-1,-i,i\}$, since $H$ is cyclic, this implies $Z(H) = H $. But $H \ne \{ 1,-1\} \cap H = \{1,-1\}$. A non-trivial example for which this is true is $G = GL_n(\mathbb{F})$ and $H = SL_n(\mathbb{F})$. $\, Z(SL_n(\mathbb{F})) = Z(GL_n(\mathbb{F}))\cap SL_n(\mathbb{F})$ = All scalar matrices with determinant $1$.

Is there a way to classify all these pair of groups ? Are there finitely examples which doesn't satisfy this ?

  • $\begingroup$ I have no idea what "classify" means here. I think the question is too broad to allow a straightforward answer. $\endgroup$ – Derek Holt Oct 16 at 7:47
  • $\begingroup$ By classify, I mean finding all possible solutions which satisfy certain properties/relations, as mentioned in solution posted by Sir Nicky Hekster. $\endgroup$ – Sabhrant Oct 16 at 7:55

This is true for example in the following cases. If (a) $G=HZ(G)$, or (b) $H \unlhd G$ and $H \cap G'=1$ (in (b) note that $H \subseteq Z(G)$). This can be generalized by using the concept of isoclinism between groups (write $\sim$), an equivalence relation on the class of groups coarser than isomorphism (see for instance my paper here). In case $G$ is finite, then your property is true whenever (a) $G \sim H$ or (b) $H \unlhd G$ and $G \sim G/H$.

  • $\begingroup$ Not sure if I understand you well, take $G=Q \times C_2$, (Q the quaternion group of order $8$), $H=Q \times \{1\}$. Then $G=HZ(G)$ and $Z(H)$ is non-trivial. $\endgroup$ – Nicky Hekster Oct 16 at 8:58
  • $\begingroup$ Yes sure, $G=HZ(G)$ implies $Z(H)=H \cap Z(G)$, that is my point. But $Z(H)$ does not need to be trivial. $\endgroup$ – Nicky Hekster Oct 16 at 9:10
  • $\begingroup$ OK, I get it. I'm deleting my comments. $\endgroup$ – Mark Bennet Oct 16 at 9:19

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