For any set X, let S_{X} be the symmetric group on
X, the group of permutations of X.

My question is: Can there be two nonempty sets X and Y with
different cardinalities, but for which S_{X} is
isomorphic to S_{Y}?

Certainly there are no *finite* examples, since the symmetric
group on n elements has n! many elements, so the finite
symmetric groups are distinguished by their size.

But one cannot make such an easy argument in the infinite
case, since the size of S_{X} is 2^{|X|},
and the exponential function in cardinal arithmetic is not
necessarily one-to-one.

Nevertheless, in some set-theoretic contexts, we can still
make the easy argument. For example, if the Generalized
Continuum Hypothesis holds, then the answer to the question
is No, for the same reason as in the finite case, since the
infinite symmetric groups will be characterized by their
size. More generally, if κ < λ implies
2^{κ} < 2^{λ} for all
cardinals, (in another words, if the exponential function
is one-to-one, a weakening of the GCH), then again
S_{κ} is not isomorphic to
S_{λ} since they have different
cardinalities. Thus, a negative answer to the question is
consistent with ZFC.

But it is known to be consistent with ZFC that
2^{κ} = 2^{λ} for some
cardinals κ < λ. In this case, we will
have two different cardinals κ < λ, whose
corresponding symmetric groups S_{κ} and
S_{λ} nevertheless have the *same* cardinality. But can we still
distinguish these groups as groups in some other (presumably more group-theoretic) manner?

The smallest instance of this phenomenon occurs under Martin's Axiom plus ¬CH, which implies 2^{ω} =
2^{ω1}. But also, if one just forces ¬CH by adding Cohen reals over a model of GCH, then again 2^{ω} =
2^{ω1}.

(I am primarily interested in what happens with AC. But if there is a curious or weird counterexample involving ¬AC, that could also be interesting.)

there are no finite examples... um, 0! = 1!, right? $\endgroup$nonemptysets X, Y. $\endgroup$5more comments