Let $T$ be a cyclic subgroup of a group $G$ such that $T$ is normal in $G$. Let $S$ be a subgroup of $T$. What can we say about whether or not $S$ is normal in $G$?
My work:
Let $T \colon = \langle a \rangle$ for some $a$ in $G$. Since $S$ is a subgroup of $T$, we can write $S$ as $S = \langle a^n \rangle$, where $n$ is the smallest positive integeer such that $a^n$ is in $S$.
Now since $T$ is normal in $G$, we can say that $gtg^{-1} \in T$ for all $t \in T$ and for all $g \in G$.
Now to determine whether or not $S$ is normal in $G$, we let $s$ be an element of $S$ and $g$ an element of $G$. Then $s = a^{kn}$ for some integer $k$.
Since $S \subset T$, we must have $gsg^{-1} \in T$. Or, $$ga^{kn}g^{-1} = a^m$$ for some integer $m$, since $T$ is generated by $a$.
Now in order for $S$ to be normal in $G$, we must have $m$ to be a multiple of $n$. So we take $m = qn + r$, where $q$, $r$ are integers such that $0 \leq r < n$. Therefore, we have $$ga^{kn}g^{-1} = a^{qn+r} = a^{qn} a^r, $$ whence $$ a^r = a^{-qn}ga^{kn}g^{-1}.$$ What next?
Looking at the generator $a$ of $T$, the normality of $T$ says that we must have
$$gag^{-1} = a^r$$
for some integer $r$ (that depends on $g$ of course, and $r$ must be coprime to the order of $T$ [if $T$ is infinite cyclic, we must have $r = \pm 1$], but we don't use that).
Then we use the fact that conjugation by an element of $G$ is an automorphism, in particular,
$$g\bigl(a^{kn}\bigr)g^{-1} = (gag^{-1})^{kn}.$$
The normality of $S$ follows from these observations.