Centre of a group and normalizers Let $G$ be a group and let $A \subset G$ be a non empty subset of $G$.Define the following subsets of $G$
$$Z(G) = \{z \in G \space | \space zx =xz \space \space \forall x \in G \}$$
$$N_G(A) = \{h \in G \space | \space hah^{-1} \in A \space \forall a \in A \} $$
$Z(G)$ is the center of the group $G$ and $N_G(A)$ is the normalizer of $A$ in $G$.
I now need to prove that $Z(G)$ is a subgroup of $N_G(A)$ and that $N_G(A)$ is a subgroup of $G$
Well I began by proving that $N_G(A)$ is a subgroup of $G$.
(1) let $a \in A$, now $e \in G$ and $eae^{-1} = a$ and so $e \in N_G(A)$ and so $N_G(A)$ is non empty
(2) let $h_1,h_2\in N_G(A)$ so that means that for any $a \in A$ we have $h_{1}ah_{1}^{-1} \in A$ and $h_2ah_2^{-1} \in A$, now consider $h_1h_2$,    we have $h_1h_2a(h_1h_2)^{-1} = h_1h_2ah_2^{-1}h_1^{-1}$ and since $h_2ah_2^{-1} \in A$ then we can let $y=h_2ah_2^{-1}$ and so $h_1h_2ah_2^{-1}h_1^{-1} = h_1yh_1^{-1}$ and since $y \in A$ then $h_1yh_1^{-1} \in A$ and hence $h_1h_2 \in N_G(A)$
(3) If $h \in N_G(A)$ then $h^{-1}hah^{-1}(h^{-1})^{-1} = h^{-1}hah^{-1}h$ and then how to argue that $h^{-1} \in N_G(A)$ ?
Ok in order to prove that $Z(G)$ is a subgroup of $N_G(A)$ 
I first showed that $Z(G)$ is a subgroup of $G$ and then i proved that $Z(G)$ is a subset of $N_G(A)$ is that valid ?
(1) To prove that $Z(G)$ is a subgroup, i showd that $e \in Z(G)$ since $ex=xe \space \forall x \in G$
Then if, $z_1,z_2 \in G$ then $z_1x=xz_1$ and $z_2x=xz_2 \space \forall x \in G$ and since $z_2x \in G$ and $xz_1 \in G$ so $z_1z_2x = xz_1z_2 \space \forall x \in G$
Now if $z \in Z(G)$ then $z^{-1} \in Z(G)$ but why ?
As you can see i have problem with proving the inverse exists in both questions
And to prove it's a subset , we simply see that 
$$Z(G) = \{z \in G \space | \space zx =xz \space \space \forall x \in G \}$$
and so since $A \subset G$ then in particular for every $z \in Z(G)$, we have 
$za=az \space \forall a \in A$ and $za=az$ implies that $zaz^{-1}= a$ and so $Z(G) \subset N_G(A)$ and hence $Z(G)$ is a subgroup of $N_G(A)$
I need suggestions to prove the inverses 
 A: Consider the following proposition:
Let $G$ be a group.  Then $H \subseteq G$ is a subgroup if and only if $e \in H$ and for any $x,y \in H, \; xy^{-1} \in H$.
$\Longrightarrow$
Suppose $H$ is a subgroup.  Then if $x,y \in H$ then clearly $y^{-1} \in H$ so $xy^{-1} \in H$.
$\Longleftarrow$
Let $H$ contain $e$ and for any $x,y \in H$ we have $xy^{-1} \in H$.  If $x \in H$ is any element just consider the product $ex^{-1} = x^{-1} \in H$, hence $H$ is closed under inverses.  Since $H$ is closed under inverses we have that given that $y \in H$ then $y^{-1} \in H$ hence we know that $x(y^{-1})^{-1} = xy \in H$.  This proves $H$ is a subgroup.  $\Box$
Can you use this proposition to make your proofs above easier?  Also, in your query about $Z(G)$, the answer is yes: If $H$ and $K$ are subgroups of $G$ and $K \subseteq H$, then $K$ is also a subgroup of $H$.
A: I can give you a hint for the center: use the fact that $g = ge = gzz^{-1}$.
For the normalizer I think there is an issue with the definition. On Wikipedia (I have never personally tried to normalize anything that wasn't a subgroup) I see it defined using $zAz^{-1} = A$. If $A$ is infinite then I think this can be stronger than your condition: multiplication by $2$ is an automorphism of the additive group $\mathbf{R}$ taking the subset $[1,\infty)$ into itself, but the inverse automorphism does not do this. Using a semidirect product you can turn this into an example of the form $\mathbf R \rtimes \mathbf Z$.
Someone smarter than I can probably come up with a more natural example. Or I'm wrong!
