Infinite Abelian subgroup of infinite non Abelian group example My thought is that we may take GL(2,F) as the group and this is obviously infinite and non abelian since matrix multiplication does not commute. Then I thought that if we make $\langle g\rangle$, for some $g$ in $\mathrm{GL}(2,F)$, which will be cyclic and hence Abelian, for instance:
$ g= 
\bigg[
\begin{matrix}
1&amp;0\\0&amp;2 
\end{matrix}
\bigg]
$. Then $g^n$ will be in the form $ g^n= 
\bigg[
\begin{matrix}
1&amp;0\\0&amp;2^n 
\end{matrix}
\bigg]
$. This is obviously infinite since $g^n=e \Leftrightarrow n = 0$.
Would this example work? Much thanks in advance!
The simplest example is $G=\mathbb Z \times S_3$ and $H=\mathbb Z$.
Yes it works if you take $F$ to be an infinite field for example.

Although, as was pointed out by others even in this case you'll have to make some assumptions on $F$ to get your particular example working.

I guess it'd be more natural to consider the subset of all diagonal submatrices. It certainly is a subgroup as $\mathrm{diag}(x,y)^{-1} = \mathrm{diag}(x^{-1}, y^{-1})$.

This subgroup is isomorphic to $F^\times \oplus F^\times$, which is abelian and infinite if $F$ is.