Question

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.)

Answer 1

According to the Schreier–Ulam–Baer theorem, the nontrivial normal subgroups of $S(X)$ are (i) the subgroup $S_\mathrm{fin}(X)$ of permutations of $X$ of finite support, (ii) the subgroup $A_\mathrm{fin}(X)$ of $S_\mathrm{fin}(X)$ of even permutations, and (iii) for each cardinal $\kappa$ the subgroups $S_{<\beta}(X)$ and $S_{\leq\beta}(X)$ of permutations which move strictly less than $\beta$ points and at most $\beta$ points, respectively.

Since, as you said, a cardinal is determined by the order type of the set of cardinals below it, looking at the lattice of normal subgroups of $S(X)$, then, lets you guess the cardinal of $X$.

Answer 2

In an ancient paper with Saharon Shelah, I proved that if κ < λ, then Sym(λ) does not embed into Sym(κ). The proof is based on results in an even more ancient paper with John Dixon and Peter Neumann. The relevant papers are:

Saharon Shelah and Simon Thomas, Implausible subgroups of infinite symmetric groups. Bull. London Math. Soc. 20 (1988), no. 4, 313--318.

John D. Dixon, Peter M. Neumann and Simon Thomas, Subgroups of small index in infinite symmetric groups. Bull. London Math. Soc. 18 (1986), no. 6, 580--586.

Answer 3

See exercises 4.6.5 - 4.6.8 in Dummit & Foote, 3rd edition. In particular, the Schreier-Ulam theorem is not needed.

Steve