To overcome the problem of an uncountable set of people speaking one after the other, note that in the original puzzle, when person i needs to speak, this person has already heard the guesses of all those preceding him. These guesses are all known to be correct, except for the first person’s guess. Furthermore, person i can see the colors of all subsequent people. Thus the information person i uses is the guess of the first person and the hat colors of all participants except for himself (and the first person).

So the generalized setting is the following. One person is designated as “first”. Everybody gets to see the hats of all others but not their own. (Possibly, they also don’t get to see the first person’s hat; this makes no difference.) The first person then makes a guess heard by all. Thereafter no communication occurs; all silently write down their guesses.

Now, take a subset of R of cardinality aleph1.
The free group generated by it (with the regular addition or multiplication in R) is a subgroup of R with cardinality aleph 1.

What about the second part? Given a cardinal A, is there a group of cardinality A?
More specifically, I wonder if there is a subgroup of R of cardinality aleph 1.
Any thoughts?

I don’t understand your phrase “any group + some more”.

For each cardinal A, if you have a group G of cardinality A, then the solution works. I mean that you don’t need anything more than a group. Disagree?

On another note, it remains to show that for each cardinal A, there is a group of cardinality A.

And another little comment (about the original riddle and the new one): the number of possible hat colors could be even more than Aleph0. The hats could be organized in any set with a “+” and a “-” operations (any group + some more). Once a man knows the “Sum” of all the hats in front of him B, and the “Sum” of all the hats in front him including him A, he can calculate the “Difference” (A – B) giving his hat color. The algorithm will still work. Ofcourse you would like the men to be able to say-out-loud the color, so that’s quite reduces the options to Aleph0 or finie sets.

—– SPOILER – possible solution below —–

We’ve solved the riddle if we can find two sets of infinite bit-vectors, called “Zero” and “One”, such that:
1. Zero and One are disjoint and their union contains all possible infinite bit-vectors.
2. For each vector in Zero, flipping one bit produces a vector in One.
3. For each vector in One, flipping one bit produces a vector in Zero.

If we can find two such sets, then the infinite men will agree on these sets before the task begins.
When it begins, the first guy sees the vector of all the hats ahead of him in line. He yells “white” if the vector is in Zero, or “black” if the vector is in One.
Every other guy knows all the bits in the vector except his own (the previous by hearing, the next by seeing). Also, he knows whether the vector is in Zero or in One. Due to properties 2 and 3 of the sets Zero and One, this is enough to determine the missing bit.

Now, to prove the existence of these 2 sets:
At first this seemed grim to me because both sets are of the same cardinality as R: infinite bit vectors are isomorphic to real numbers between 0 and 1, and both sets contain infinite vectors that are isomorphic to irrational numbers.
However, maybe the axiom of choice saves us.
For each infinite bit vector v, define Bubble(v) as the set of all vectors reachable from v by flipping a finite number of bits.
Obviously there are infinite disjoint bubbles. Using AC, we choose one vector to represent each bubble, and define Zero_0 to be the set of all representatives chosen.
Then we continue in the following manner:
One_0 = {all vectors one bit away from vectors in Zero_0}
Zero_1 = {all vectors one bit away from vectors in One_0}
One_1 = {all vectors one bit away from vectors in Zero_1}
…
Zero_k = {all vectors one bit away from vectors in One_k-1}
One_k = {all vectors one bit away from vectors in Zero_k}
…
And finally:
Zero = union Zero_0, Zero_1, …
One = union One_0, One_1, …

Unless I’m wrong, I think it can be shown (though I don’t feel like formulating a proof right now) that these sets fulfill the requirements I set above.