Infinite Abelian subgroup of infinite non Abelian group example

The simplest example is $G=\mathbb Z \times S_3$ and $H=\mathbb Z$.

This example works indeed, if $F$ is infinite and $2^n\neq1$ in $F$ for all non-zero $n\in\Bbb{Z}$. This is satisfied for obvious candidates for $F$ such as $\Bbb{R}$, $\Bbb{C}$ and $\Bbb{Q}$, but fails for other candidates such as the finite fields $\Bbb{F}_q$, but also infinite fields of positive characteristic such as $\Bbb{F}_p(T)$.

Assuming $F$ is a field, the condition that $2^n\neq1$ for all non-zero $n\in\Bbb{Z}$ is equivalent to $\operatorname{char}F=0$, from which it follows that $F$ is infinite. So your example works if and only if $\operatorname{char}F=0$.

Assuming that $\Bbb F$ has characteristic $0,$ that definitely works. Nicely done!

It also allows you to prove an inclusion $\Bbb Z\hookrightarrow GL(2,\Bbb F).$