Are there groups $G$, whose order is uncountable, such that the order of $G/[G,G]$ is countable? 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)?
 A: 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 $\Omega.$
A: 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.
A: Another class (even not a set) of examples of uncountable perfect groups:
Let's for any cardinality $\alpha$ define $S_\alpha$ as the group of all permutations of a set of cardinality $\alpha$. 
On one hand, it is a known fact that:
$$|S_\alpha| = 2^\alpha$$
On the other hand, the Theorem 6 of "Some remarks on commutators" by Oystein Ore states:

If $\alpha \geq \aleph_0$ then $\forall g \in S_\alpha \exists h, k \in S_\alpha$, such that $g =[h, k]$

From that it follows that $S_\alpha' = S_\alpha$ and thus $S_\alpha$ has trivial abelianizaiton.
