Show $Z(H)Z(G)=(HZ(G))\cap C_G(H)$ for $H\le G$ Suppose that $H$ is a subgroup of $G$, then I would like to show that the following follows:
$$Z(H)Z(G)=(HZ(G))\cap C_G(H)$$
where $C_G(H)=\{ x\in G \mid \forall h\in H: xh=hx\}$ and $Z(G)=\{ z\in G\mid \forall g\in G, zg=gz\}$.
What I showed thusfar is the following:
Let $y\in Z(H)Z(G)$ so there exist $z_1\in Z(H), z_2\in Z(G)$ such that: $y=z_1z_2$, we get that $y\in C_G(H)$ since $\forall h\in H: \ \ hz_1z_2=z_1hz_2=hz_1z_2$, so $yh=hy$.
Now I find it difficult to show that: $y\in HZ(G)$.
Here's how I started my proof:
$$y=z_1z_2=hz_1h^{-1}z_2=h(z_1h^{-1}z_2)$$
where $h\in H$ is some arbitrary element of the subgroup $H$. So, I need to show that: $z_1h^{-1}z_2\in Z(G)$, which is where I am stuck? I need to show that for each $g\in G$ we have: $gz_1h^{-1}z_2=z_1h^{-1}z_2g$, how to show that? something trivial I am missing here?
Thanks!
BTW, is this proof of one direction is reversible, i.e instead of one direction it's actually bi-directional, or is the other direction different than this direction?

You could almost do this hands down if you are familiar with Dedekind's Modular Law: if $A,B,C$ are subgroups of a group $G$ with $A \subseteq B$ then (as sets) $(B \cap C)A = B \cap CA$. Hence, since, $Z(G) \subseteq C_G(H)$ we get
$$Z(H)Z(G)=C_H(H)Z(G)=(H \cap C_G(H))Z(G)=HZ(G) \cap C_G(H).$$
