Prove that the cyclic subgroup $\langle a \rangle$ of a group $G$ is normal if and only if for each $g \in G$, $ga = a^k g$ for some $k \in \mathbb{Z}$.
Suppose $\langle a \rangle$ is normal in $G$. Then $\langle a \rangle g = g \langle a \rangle$, for all $g \in G$. This implies $a^k g \in g \langle a \rangle$ for some $k \in \mathbb{Z}$.
(How can I make the connection from this point onward to the conclusion of the proof in the forward direction?)
Conversely, suppose $ga = a^k g$. Then $ga \in \langle a \rangle g$ and so $g \langle a \rangle \subseteq \langle a \rangle g$. Now $a^k g = ga$ implies $a^k g \in g \langle a \rangle$, and so $\langle a \rangle g\subseteq g \langle a \rangle$. We finally get $\langle a \rangle g = g \langle a \rangle$.
Best Answer
A subgroup $U\le G$ is normal if and only if $gU=Ug$ for all $g\in G$. This translates elementwise to the statement
This is the inclusion $gU\subseteq Ug$ for all $g$, but $g^{-1}U\subseteq Ug^{-1}$ already implies $$ Ug = g(g^{-1}U)g \subseteq g(Ug^{-1})g = gU, $$ so the statement is equivalent to $gU=Ug$ for all $g\in G$.
Now let $U=\langle a\rangle = \{\,a^k \mid k\in\mathbb Z\,\}$. Now $U$ is normal if and only if
Assume $U$ is normal, then we get $ga=a^kg$ for some $g$ by choosing $l=1$ in the statement above.
For the converse, assume that $ga=a^kg$ for some $k$. Given any $l\in\mathbb Z$ we get $$ ga^l = (ga^l g^{-1}) g = (gag^{-1})^l g = (a^k g g^{-1})^l g = a^{kl} g, $$ so the statement above is true and therefore $U$ is normal.