Two questions:

  1. Suppose a and b are two uncountable cardinals. Consider the symmetric groups on sets of sizes a and b respectively (the symmetric group on a set is the group of all bijections from the set to itself, under composition). Consider the first-order theories of these as "pure groups" (i.e., just the group structure, no additional information). Are these elementarily equivalent? (does the answer change if we allow one of a or b to be the countable cardinal?)

  2. Suppose $a_1$, $a_2$, $b_1$ and $b_2$ are uncountable cardinals with $a_1 < a_2$ and $b_1 < b_2$. Consider the symmetric group on a set of size $a_2$ and the subgroup of those bijections that have support of size at most $a_1$. Consider the pure theory of this group-subgroup pair (i.e., the pure theory of the group along with a membership predicate for the subgroup). Similarly, for $b_1$ and $b_2$. Are these two pure theories elementarily equivalent? Does the answer change if, instead, we look at the subgroup of those bijections that have support of size strictly less than $a_1$? Does the answer change if we allow the countable cardinal? (The support of a bijection is the set of elements that are moved).

By the Baer-Schreier-Ulam theorem, the only normal subgroups of symmetric groups on infinite sets are the subgroups comprising bijections with support of size strictly less than a for some infinite cardinal a or the subgroups comprising bijections with support of size less than or equal to a for some infinite cardinal a, plus the trivial subgroup and the finitary alternating group. All these are also characteristic subgroups.

If (2) is true, this would give examples of distinct characteristic subgroups of the same group that are elementarily equivalently embedded as subgroups (i.e., there is no first-order sentence true for one subgroup that is not true for the other).


For each ordinal $\alpha < \omega^\omega$ the symmetric group on $\aleph_{\alpha}$ is first order definable in the class of all symmetric groups; i.e. there is a sentence in the first order language of groups that is true in $Sym(A)$ iff $|A| = \aleph_\alpha$. See Mckenzie - On elementary types of symmetric groups, Algebra Universalis 1 (1971), 13--20.

Shelah provides an interesting necessary and sufficient condition for the elementary equivalence of $Sym(\aleph_\alpha)$ and $Sym(\aleph_\beta)$ - First order theory of permutation groups, Israel J. Math. 14 (1973), 149–162; corrections: ibid. 15 (1973), 437–44.

| cite | improve this answer | |
  • $\begingroup$ Amazing! Can you tell us the Mckenzie sentences, or Shelah's necessary and sufficient condition? $\endgroup$ – Joel David Hamkins Jan 10 '10 at 1:32
  • $\begingroup$ Shelah's result can be found here (pages 414-15) iop.org/EJ/article/0036-0279/60/3/R01/… I haven't read Mckenzie's paper yet. $\endgroup$ – Ashutosh Jan 10 '10 at 1:55
  • $\begingroup$ Amazing! There is a third paper by A. G. Pinus [Algebra Universalis 3 (1973), 59--66] which extends to $\alpha < \omega_1^{CK}$, if I understood correctly. From the summary, the trick (due to Ershov) is to interpret the two-type structure $(M,Sym(M);R)$ inside $Sym(M)$, where $R(s,x,y)$ says $s(x) = y$. Then, since there are plenty of bijections around, you can accurately compare the size of various subsets of $M$. $\endgroup$ – François G. Dorais Jan 10 '10 at 2:09
  • $\begingroup$ I see. That is very interesting. $\endgroup$ – Joel David Hamkins Jan 10 '10 at 2:17

For question 1, it must be true for huge numbers of cardinal pairs, for the simple reason that there are only continuum many first order theories in a countable language, but there are more than continuum many uncountable cardinals. Thus, in fact there is a proper class of cardinals serving as positive examples of your phenomenon.

The same idea answers part of question 2. We have essentially a map from pairs of cardinals to the corresponding theory, and since there are again only continuum many theories, there must be a proper class of pairs of cardinals getting the same theory.

I am expecting that they are all elementary equivalent.

[Edit: my expectation is apparently refuted by Mckenzie and Shelah.]

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.