# Let $G$ be a group, and $H \le G$. For what pairs $(G,H)$ is $Z(H)=Z(G)\cap H$ true?

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 $$H = = \{1,-1,-i,i\}$$, 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 ?

• I have no idea what "classify" means here. I think the question is too broad to allow a straightforward answer. Commented Oct 16, 2019 at 7:47
• By classify, I mean finding all possible solutions which satisfy certain properties/relations, as mentioned in solution posted by Sir Nicky Hekster.
– Sam
Commented Oct 16, 2019 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 $$G \sim G/H$$.
• 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. Commented Oct 16, 2019 at 8:58
• 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. Commented Oct 16, 2019 at 9:10