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Are there groups $G$, whose order is uncountable, such that the order of $G/[G,G]$ is countable?

I am mostly concerned with looking at the groups in terms of generators and relations, so this can be rephrased to be: are there groups with uncountable many generators (needed) but the abelianized group can be presented with countably many generators? If so are there uncountable groups that after abelinaization are finitely generated?

Is there a simple way to construct groups (if they exist)?

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What you ask in your question's title and in its body are two different things: uncountable groups s.t. their abelianization is countable is one thing, groups with uncountable generators but with countable abelianization is another one. Which one do you want? – DonAntonio Dec 19 '13 at 21:01
@DonAntonio Oh, wouldn't an uncountable group have uncountably many generators? I guess what is the difference? I didn't realize that they were that different. I guess you could have uncountable many generators but have a countable groups which is why I put that "needed" in parenthesis. If that is the problem then I will edit to be more clear that I am talking about uncountable groups. – Paul Plummer Dec 19 '13 at 21:09
Yes, of course $\;G\;$ uncountable $\;\implies\;$ uncountable generators, but this has nothing to do necessarily with uncountable number of such groups, as the title mentions. – DonAntonio Dec 19 '13 at 21:13
Oh I am talking about a group being uncountable, the order of the group, sorry. – Paul Plummer Dec 19 '13 at 21:14
Ok, now I got you....hopefully. :0 – DonAntonio Dec 19 '13 at 21:15
up vote 5 down vote accepted

Yes, in the group $PSL(n, \mathbb{C})$ every element is a commutator, so the quotient has cardinality $1.$ A weirder example is the permutation group $S_\Omega$ of an infinite set $\Omega.$

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In a similar vein: take an uncountable product of perfect groups. – Nicky Hekster Dec 19 '13 at 21:20
Also the alternating group $A_{\kappa}$, where $\kappa$ is any infinite cardinal. Such groups are simple, and so work. – user1729 Dec 20 '13 at 9:28
@user1729 What is the alternating group of an infinite cardinal? Do you mean permutation groups whose support is finite (which is what you need to define parity)? That's fine, but different from the kind of groups I meant, so this is a completely different example... – Igor Rivin Dec 20 '13 at 13:58
Yeah, even permutations with finite support. I wasn't sure if they counted as related to $S_{\Omega}$ or not (in fact, is $\operatorname{PSL}(n, \mathbb{C})$ definitely not isomorphic to an alternating group?). I'll post them as another answer then. – user1729 Dec 20 '13 at 14:08

As I said in my comment to Igor Rivin's answer, if $\kappa$ is an infinite cardinal then the group $A_{\kappa}$ of even permutations with finite support is simple. Therefore, the abelianisation of $A_{\kappa}$ is trivial.

In fact, as $A_{\kappa}$ has cardinality $\kappa$ this yields an example for every possible cardinality.

To see that $A_{\kappa}$ is simple, apply the following lemma to the fact that $A_n$ is simple for $n\geq 5$. (See page 73 of D.J.S. Robinson's book A course in the theory of groups.)

Lemma: If $G$ is the union of a chain of simple groups, so $G=\cup H_i$ where $H_0\leq H_2\leq H_3\leq\ldots$ and each $H_i$ is simple, then $G$ is simple.

Proof: Suppose $1\neq N\unlhd G$. We shall prove that $N=G$. Because $G$ is the union of a chain of simple groups there exists some simple group $H_i$ with $N\cap H_i\neq 1$, and then $N\cap H_j\neq 1$ for all $j>i$. But then $N\cap H_j\lhd H_j$ so $H_j\leq N$ for all $j>i$. We thus conclude that $N=G$, as required.

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