Prove that the cyclic subgroup $\langle a\rangle$ of a group $G$ is normal if and only if for each $g\in G$, $ga=a^kg$ for some $k\in\Bbb{Z}$. 
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 $\langle a \rangle g = g \langle a \rangle$.
 A: A subgroup $U\le G$ is normal if and only if $gU=Ug$ for all $g\in G$. This translates elementwise to the statement

For every $g\in G$ and $u\in H$ there is some $u'\in U$ such that $gu=u'g$.

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

For every $g\in G$ and $l\in\mathbb Z$ there is some $k\in\mathbb Z$ such that $ga^l=a^k g$.

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.
A: You have to prove that $ga=a^k g$ for some $k$. You have shown that $g\langle a\rangle=\langle a\rangle g$. Since $ga$ is an element of $g\langle a\rangle$, you know that $ga$ is an element of $\langle a\rangle g$. This means that there exists some element $h\in\langle a\rangle$ so that $a=hg$. Can you prove that $h=a^k$ for some $k$?
For your converse, your proof is sloppy. Sure, $a^kg = ga$ implies that $a^kg\in g\langle a\rangle$. However, this does not (yet) prove that $\langle a\rangle g \subseteq g\langle a\rangle$. This is because to prove this, you must prove that every element of $\langle a\rangle g$ is an element of $g\langle a\rangle$. All you did was prove that $a^kg$ (for some $k$) is an element of $g\langle a\rangle$.
In fact, I am not even convinced at your proof that $g\langle a\rangle\subseteq\langle a\rangle g$. You say that this follows from $ga\in\langle a\rangle g$, however you must prove that $ga^n\in\langle a\rangle g$ for every $n$, which you have, at least by my oppinion, not done (at least not rigorously).
