Infinite Abelian subgroup of infinite non Abelian group example

Question

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&0\\0&2 \end{matrix} \bigg] $. Then $g^n$ will be in the form $ g^n= \bigg[ \begin{matrix} 1&0\\0&2^n \end{matrix} \bigg] $. This is obviously infinite since $g^n=e \Leftrightarrow n = 0$. Would this example work? Much thanks in advance!

Answer 1

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 $\operatorname{char}F=0$.

Answer 2

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.

Answer 3

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

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

Answer 4

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

Answer 5

A simple example: let $G = S(\mathbb Z)$, the group of all permutations of the integers. Let $A$ be the subgroup generated by the transpositions $\{(n,n+1) |\, n \in \mathbb Z, n \, \rm{ even} \}$. Since the generating transpositions are all pairwise disjoint, they trivially commute with each other.