If $H$ is a cyclic subgroup of $G$ and $H$ is normal in $G$, then every subgoup of $H$ is normal in $G$. Exercise 11, page 45 from Hungerford's book Algebra.

If $H$ is a cyclic subgroup of $G$ and $H$ is normal in $G$, then every
  subgroup of $H$ is normal in $G$.

I am trying to show that $a^{-1}Ka\subset K$, but I got stuck. What I am supposed to do now?
Thanks for your kindly help.
 A: Suppose $H = \langle h \rangle$ is normal in $G$ and that $K$ is a subgroup of $H$. Any subgroup of a cyclic group is cyclic, so $K = \langle h^d \rangle$ for some integer $d$. 
Let $g \in G$. Since $H$ is normal, $g^{-1}hg = h^i$ for some integer $i$. Then for any integer $k$ you get $g^{-1}(h^d)^kg = (g^{-1}hg)^{dk} = (h^i)^{dk} = (h^d)^{ik}$. This shows that for any $k \in K$, the element $g^{-1}kg$ is in $K$. Therefore $K$ is normal.
A: since $H$ is normal in $G$ you get $a^{-1}Ka \subset H$, for all $a\in G$. Now use the fact that $H$ is cyclic (there is only one subgroup of $H$ such that $\dots$)
A: Here is a somewhat more general fact which seems useful enough to keep in mind:

If $G$ is a group, $H$ is a normal subgroup of $G$ and $K$ is a characteristic subgroup of $H$, then $K$ is a normal subgroup of $G$.

The proof is almost immediate if you know the definitions: for any $x \in G$, since $H$ is normal in $G$, conjugation by $H$ induces an automorphism $\varphi_x$ of $H$, but not necessarily an "inner" automorphism: i.e., if $x \notin H$, $\varphi_x$ need not be conjugation by any element of $H$.  Thus we have assumed that $K$ is just not normal but characteristic as a subgroup of $H$, i.e., stable under all automorphisms of $H$.  Done.
For much more detail, see e.g. here.
As others have pointed out, we also need to see that any subgroup of a cyclic group $H$ is characteristic.  Well, any subgroup which is the unique subgroup of its order is characteristic -- this takes care of the case in which $H$ is finite.  And any subgroup which is the unique subgroup of its index is characteristic -- this takes care of the case in which $H$ is infinite.  (Alternately, if $H \cong (\mathbb{Z},+)$, the only nontrivial automorphism is multiplication by $-1$, which evidently stabilizes all the subgroups $n \mathbb{Z}$.)
A: I'll give a try. If $H=\langle h \rangle$ , then $H$ is an abelian group and $K$ is a normal subgroup of $H$. Let $d$ the lowest positive integer such that $h^{d}\in K$. Then  $K=\langle h^{d} \rangle$ and we have $H/K=\{K,hK,\cdots,h^{d-1}K\}$. Let $g\in G$  and $k=h^{dn}\in K$. Then $gkg^{-1}K=(ghg^{-1})^{dn}K=K$. Thus $gKg^{-1}\subset K$, for all $g\in G$.
I hope that it is correct. 
A: Let $H=\langle h\rangle$. $H\unlhd G\implies\forall g\in G, ghg^{-1}=h^m$ for some $m$. All elements of $K$ looks like $h^n$. $\forall g\in G, gh^ng^{-1}=(ghg^{-1})^n=(h^m)^n=h^{mn}=(h^n)^m\in K$
A: Another proof, suppose that $H=<h>$,$m,n\in \mathbb{N}$, $q,p\in \mathbb{Z}$ such that $m=pn+q$, $q<n$.
Generally, $x^{-1}h^{m}x=h^{n} \Leftrightarrow x^{-1}h^{pn+q}x=h^{n} \Rightarrow h^{qo(h^{n})}=e \Rightarrow q=0$.
