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 $k\in\Bbb{Z}$.

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.