Show that $H \leq C_G(C_G(H))$. Show that $H \leq C_G(C_G(H))$.

We have that $C_G(H) = \left\{ g \in G | gh = hg \; \forall h \in H \right\}$. So then $C_G(C_G(H))$ is the $g$'s that commute with all $g$'s that commute with $h$. This would mean $$C_G(C_G(H)) = \left\{ g \in G | gg'h = hg'g, \; \forall g' \in G, \; \forall h \in H \right\}$$
But since $gg' \in G$ this is just some element of $G$, so doesn't this become the same as $C_G(H)$?

I feel like what I have is wrong but I am not sure why.
Pick $h\in H$. We wish to show that $hg=gh$ for any $g\in C_G(H)$. But this is tautological from the definition of $C_G(H)$, so $H\subset C_G(C_G(H))$.
in the definition $ C_G(C_G(H)) = \left\{ g \in G | gg'h = hg'g, \; \forall g' \in G, \; \forall h \in H \right\}$, is not $\forall g' \in G$, but $\forall g' \in C_G(H)$. 

I believe that the question is wrong, because: 

If $h_1\in C_G(C_G(H))$ and $h_1 \in H$, we have $$ h_1g'h_2=h_2g'h_1  ,\forall g' \in C_G(H), \forall h_2 \in H  $$
It's clear that the identity $e \in C_G(H)$, so $$h_1h_2=h_2h_1, \forall h_2 \in H $$, then $H \subset C_G(C_G(H))$ only if H is abelian. But $H \leq C_G(C_G(H))$ also can indicate relation of order.