Prove that $H$ is a normal subgroup of $G$. 
Let $G$ be a finite group, $N$ a cyclic normal subgroup of $G$, and $H$ any subgroup of $N$. Prove that $H$ is a normal subgroup of $G$.

This proof has errors probably:
Since $N$ is cyclic, let $N$ be generated by $\langle n \rangle$, for some $n \in N$. Also, $N$ is normal in $G$, so we have $gng^{-1} \in N$. Since $H \le G$, closure holds in $H$. So for some $n \in H$ (since $n \in G$), $gng^{-1} \in H$.
 A: Let $g \in G $. 
Observe that $H^{g} \subseteq N^{g} = N$.
N is finite and cyclic, so $H$ is the only subgroup of $N$ with cardinality $|H|$. So $H^{g} = H $ and hence $H$ is normal in $G$.
A: This is a special case of something far more general: a subgroup $K$ of $G$ is called characteristic (and one writes $K$ char $G$) is it is fixed by every automorphism of $G$ ($\forall \alpha \in Aut(G): \alpha(K)=K$). Because conjugation is an automorphism, every characteristic subgroup is normal, though not every normal subgroup has to be characteristic. Examples of characteristic subgroups include the commutator subgroup and the center of a group. All subgroups of a finite cyclic group are characteristic (because of the unicity of the order of each subgroup). Now, and this is the general statement, if $K$ char $N \unlhd G$, then $K \unlhd G$. $K$ does not have to be cyclic, an example, let $K=\{(1), (1 2)(34), (1 3)(2 4), (1 4)(2 3)\}$, $N=A_4$ and $G=S_4$. Observe that $A_4'=K$, so $K$ char $A_4$. And hence $K \lhd S_4$.
