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.

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.
