If a linear group $G$ has an Abelian dense subgroup $H$, then $G$ itself is abelian

Question

I am having a lot of trouble with this proving the following statement: If a linear group $G$ has an Abelian dense subgroup $H$, then $G$ itself is abelian. More specifically, the notion of a "dense subgroup" is confusing to me. I was trying to compare this idea to maybe something like $\mathbb{Q}$ being dense in $\mathbb{R}$, but that example was too trivial to really help me with the proof I am actually trying to complete. Any help is appreciated!

Answer 1

I assume that the groups are Lie group, the map $f:G\times G\rightarrow G$ defined by $f(x,y)=xyx^{-1}y^{-1}$ is continue. $f(A\times A)=Id_G$, this implies that $f^{-1}(Id_A)$ is a closed subset which contains $A\times A$ so it is $G\times G$ since $A\times A$ is dense in $G\times G$. We deduce that $G$ is a commutative group.

Answer 2

Linear groups are examples of topological groups: these are groups equipped with a topology such that the group operations (multiplication and inverse) are continuous. Density in the context of topological groups means exactly what it means in general topology - and as a good example, $(\mathbb{Q}, +)$ is a dense subgroup of $(\mathbb{R}, +)$.

The idea is that since being abelian is defined via the group operations, if it holds densely it should hold always. This can be formalized by finding a continuous function corresponding to "abelian-ness": namely, take $f(a, b)=aba^{-1}b^{-1}$ (the commutator map). Then $f(a, b)=e$ iff $ab=ba$.

So what? Well, $f$ is continuous since the group operations are continuous (exercise), and by hypothesis is equal to $e$ on a dense set. But if a continuous function is constant on a dense set, then it's just constant - this is a general fact of topology. So $f(a, b)=e$ always, which is to say the group is abelian.

This second-to-last sentence uses the further assumption that $\{e\}$ is closed; this is not necessarily true, but is true in every "naturally-occurring" topological group. In particular, every linear group has this property, as does every metrizable group more generally.

Note that the only special thing we've used about abelian-ness is that it's given by an equation: namely, $G$ is commutative iff $ab=ba$ for each $a, b\in G$. We can rewrite such an equation to get an expression which is always $e$ iff $G$ is abelian, namely $ab=ba\implies aba^{-1}b^{-1}=e$, and we get the function $f$ from this second equation. This same process more generally shows:

If an equation in the language of groups is satisfied on a dense set, then it's satisfied everywhere.

For instance, if $a^2b^4a^6b^8=e$ for all $a, b\in D$ for some dense $D\subseteq G$, then $a^2b^4a^6b^8=e$ for all $a, b\in G$.

This fails for properties more complicated than equations - e.g. consider the group $G=(\{q+z\pi: q\in\mathbb{Q}, z\in\mathbb{Z}\}, +)$. $(\mathbb{Q}, +)$ is a dense subset of $G$, and every element of $\mathbb{Q}$ is even - that is, for each $a\in\mathbb{Q}$, there is some $b\in\mathbb{Q}$ such that $a+a=b$ - but $\pi$ is not even in $G$ (since ${\pi\over 2}$ isn't in $G$). The point is that "is even" is a more complicated statement than just "commutes" - it involves a "there exists" phrase, and it turns out that quantifiers are important here.