Normalizers and Centralizers I am trying to show the following.
Let $G$ be a group, $X \subset G$ a subset.  The $centralizer$ of $X$ in $G$ is $Z_G(X) = \{g \in G | \forall x \in X: gx = xg\} $ $-$ in particular $Z(G) = Z_G(G)$ is called the centre of $G$. The $normalizer$ of $X$ in $G$ is $N_G(X) = \{g \in G \ | \ gX g^{-1} = X\}$. Fix $H < G$.
(a) Show that $N_G(X) < G$.
(b) Show that $H<N_G(H)$.
(c) Show that $Z(G)$ is a normal, abelian subgroup of $G$.

As for (a), my strategy is to show that $a,b \in N_G(X) \implies ab^{-1} \in N_G(x)$. Given $a,b \in N_G(X)$, we have $aXa^{-1} = X$ and $bXb^{-1} = X$ which $\implies$ (implies that)  $aXa^{-1} = bXb^{-1}$ $\implies$ $ab^{-1}X(ab^{-1})^{-1} \implies ab^{-1} \in N_G(X)$.
For (b), if $a \in H$ then $\forall h \in H$, $hah^{-1} \in H$ since $H<G$. Thus $hHh^{-1} \subseteq H$. Further, if $a \in H$ then we have $a = h(h^{-1}ah)h^{-1}$ so $H \subseteq hHh^{-1}$. Therefore $H = hHh^{-1}$. Thus $h \in N_G(H)$ and $H \subseteq N_G(H)$ so $H<N_G(H)$.
For (c), $Z(G) = Z_G(G) = \{g \in G | \forall x \in G: gx = xg\}$. So for any $a,b \in Z(G)$ we obviously have $ab = ba$ since $a,b \in G$ so $Z(G)$ is abelian. It is normal since $g \in Z(G) \implies gx = xg \implies gxg^{-1} = xgg^{-1} = x \implies Z(G) \trianglelefteq G$. 
Is my reasoning correct?
 A: Point (c): You didn't show $Z(G)$ is a subgroup. Furthermore your translation of the normality of $Z(G)$ is wrong: it only translates abelianity. What you should have proved is this:
If $g\in Z(G)$, for any $x\in G$, $xgx^{-1}\in Z(G)$.
This indeed is true, since $g$ commutes with every element of $G$: $\;xgx^{-1}=xx^{-1}g=g$.
This means inner automorphisms of $G$ induce identity on $Z(G)$.
A: About your first answer :

As for (a), my strategy is to show that $a,b \in N_G(X) \implies ab^{-1} \in N_G(x)$. Given $a,b \in N_G(X)$, we have $aXa^{-1} = X$ and $bXb^{-1} = X$ which $\implies$ (implies that)  $aXa^{-1} = bXb^{-1}$ $\implies$ $ab^{-1}X(ab^{-1})^{-1} \implies ab^{-1} \in N_G(X)$.

I cannot see why $aXa^{-1} = bXb^{-1}$ $\implies$ $ab^{-1}X(ab^{-1})^{-1}$ from what you wrote. You should argue that $bXb^{-1}=X\Rightarrow X=b^{-1}Xb$ (by conjugating the former equality by $b^{-1}$) and then $aXa^{-1}=ab^{-1}Xba^{-1}$ by conjugating $X=b^{-1}Xb$ by $a$. Finally $X=aXa^{-1}=ab^{-1}Xba^{-1}$ since $a$ normalizes $X$.
The other statements look OK to me.
