Let G be a group and $a\in Z(G)$. Prove that $\langle a\rangle$ is a normal subgroup of G. 
Let $G$ be a group and $a\in Z(G)$, where
  $$
Z(G) =\lbrace x \in G\mid xg=gx~~ \forall g\in G\}.$$ Prove that $H=\langle a\rangle$ is a normal subgroup of $G$.

To show that $H$ is normal in $G$, I have to prove:


*

*$H$ is non-empty.

*$H$ is a subgroup of $G$.

*$H$ is normal in $G$.


Last part 3. Let $a\in H$ and $x\in  G$ then $xax^{-1}=axx^{-1}=a$ (the second last equality holds since, $a\in Z(G)$ then $ag=ga$, here $g=x^{-1}\in G$).
If last part is correct, please help me to show the remaining.
 A: $g=a^{-1}$ is a particular and this does not proves that $<a>$ is normal subgroup of $G$.
Solution : for showing $<a>$ is normal subgroup of $G$ we need to show that  $gag^{-1}\in <a>~~\forall g\in G$.
Now since $a\in Z(G)~\implies ag=ga~~\forall g\in G$
therefore $gag^{-1}=(ga)g^{-1}=(ag)g^{-1}=a(gg^{-1}=a\in <a>$
$\implies gag^{-1}\in <a>~~\forall g\in G$
Therefore $<a>$ is normal subgroup of $G$. 
A: That $\langle a \rangle$ is a nonempty subgroup of $G$ is clear, because $a \in \langle a \rangle$, and by definition $\langle a \rangle$ is the subgroup generated by $a$.
Your argument that $xax^{-1} = axx^{-1} = a$, shows that $xax^{-1} \in \langle a \rangle$, but this is not sufficient to prove that $\langle a \rangle$ is normal. For that, we must show that $xyx^{-1} \in \langle a \rangle$, where $y$ is any element of $\langle a \rangle$. We can do this easily: observe that $Z(G)$ is a subgroup containing $a$, and $\langle a \rangle$ is the intersection of all subgroups containing $a$ (i.e., the smallest subgroup containing $a$), hence $\langle a \rangle \leq Z(G)$. In particular, an arbitrary element $y \in \langle a \rangle$ is in $Z(G)$. So we can argue as before: $xyx^{-1} = yxx^{-1} = y \in \langle a \rangle$.
