Which of the following statements of group theory are true? Let $G$ be a group. Which of the following statements are true?
$1.$ The normalizer of a subgroup of $G $ is a normal subgroup of $G.$
$2.$ The centre of $G$ is a normal subgroup of $G.$
$3.$ If $H$ is a normal subgroup of G and is of order $2$, then H is contained in the centre of G.
My attempt:
For option $2)$ is true.
Let $x\in Z(G)$ (center of $G$).
Then for any $g\in G$, $gxg^{-1}=gg^{-1}x=x\in Z(G)$.
This proves $Z(G)$ is a normal subgroup.
I’m confused about option $1)$ and option $3)$. Are there any counterexamples? 
 A: For your first statement: 
Consider $G=S_3$ and $H=\langle (12) \rangle$.
Since $H\leq N_G(H)\leq G$, by Lagrange's Theorem, $N_G(H)=H$ or $G$.
But since $H$ is not normal, $N_G(H)=H$ which is also not normal.
A: For the first argument and as a bit difficult example than @Alan's, consider $G=D_{10}$. It is as $$\{e, a, b, ab, b^2, ab^2, b^3, ab^3, b^4, ab^4\}$$ of order $10$. For the subgroup $H=\{e, ab^3\}$ of order $2$ we can check $H=N_{G}(H)\neq G$. So the normalizer could't be a normal subgroup in $G$. Otherwise, for this example $G=N_G(N_G(H))=N_G(H)$ which violets the latter fact.
A: Not as easy an answer as is probably expected, but whenever normal subgroups are in a true / false question non-abelian simple groups are good place to start, so for statement 1:
Take any maximal subgroup $M$ of a non-abelian simple group $G$. Since $G$ is simple and $M\trianglelefteq N_G(M)\le G$ we must have $M=N_G(M)$.
A: Ad 3. Let $H \unlhd G$ and $|H|=2$. Observe that since $H$ is normal, $N_G(H)=G$. In general $N_G(H)/C_G(H)$ embeds homomorphically in $Aut(H)$ (this is called the "N/C theorem"). But $H \cong C_2$ so, $Aut(H)=\{1\}$. Hence $G=C_G(H)$, that is $H \subseteq Z(G)$.
