I want to show that given that $H \leq G$ that we also have $H \leq C_G(C_G(H)).$
My attempt :
i think the best way to do this is break it down into stages so :
Stage 1) showing that H is a subset of $C_G(C_G(H))$
$C_G(C_G(H)):=\{g\in G |cg=gc, \forall c \in C_G(H)\}$.
which is the set containing all the elements of G which commute with all the elements of $C_G(H)$. But $C_G(H)$ is the set set which contains all the elements of G which commute with all elements of H. Which means that $C_G(C_G(H)$ is the set containing all the elements of G which commute with the elements of G that commute with all the elements of h.
well h satisfies the property of of commuting with h which commutes with h. therefore H must be contained in $C_G(C_G(H))$.
Stage 2) showing that for all $a,b \in H,ab^{-1}\in H$, Although I'm not sure how to proceed from here. Any suggestions ?
I think maybe we could say that as all elements in h are of the form
(hch^{-1}gh^{-1}ch)(h^{-1}c^{-1}h g^{-1}hc^{-1}h^{-1}
then by cancelation we get that this is equal to the identity which is in H as it is a subgroup meaning that it is a subgroup of $C_GC_G(H)$
First we prove that $C_G(C_G(H)) \leq G$. Well We know that $C_G(H)\leq G \Rightarrow C_G(C_G(H))\leq G$.
We already know that $H\leq G$
So now all we need to show is that
$H\subset C_G(C_G(H))$
Well let $g\in C_G(C_G(H)$
$\Rightarrow g=(hch^{-1})g(hch^{-1})^{-1}=(hch^{-1})g(h^{-1}c^{-1}h)$.
Now suppose that g=h , now this equation is obviously true
$hch^{-1}hh^{-1}c^{-1}h=hch^{-1}c^{-1}h=hc(hc)^{-1}h=h.$ and so the condition to be a member of $C_G(C_G(H))$ is met by h therefore $H\subset C_G(C_G(H))$.
Now we use the fact that $|H|<|C_G(C_G(H))|$ and $H,C_G(C_G(H))\leq G$ to assume that $H\leq C_G(C_G(H))$