Order of a profinite group need not determine order as a group in the sense of cardinality of underlying set

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Statement

It is possible to have two profinite groups G_1 and G_2 that have the same order as each other in the sense of order of a profinite group (note that both order numbers are supernatural numbers), but that have different orders from each other in the sense of the cardinality of the underlying set.

Proof

Fix a nontrivial finite group K. Pick two infinite cardinals \alpha_1, \alpha_2 such that the power cardinals of \alpha_1,\alpha_2 are not equal. Now, consider the external direct powers (repeated unrestricted external direct product of K with itself): G_1 = K^{\alpha_1} and G_2 = K^{\alpha_2}, both equipped with the product topology from the discrete topology on K. We note that:

  • For both groups, the order in the sense of a profinite group is as follows: all primes that divide the order of K occur with a power of \infty, but no other primes occur. In particular, the order of G_1 equals the order of G_2 in the sense of order as a profinite group, where the equality is as equality of supernatural numbers.
  • The cardinality of the underlying set of G_1 is |K|^{\alpha_1}, which is the power cardinal of \alpha_1 (assuming the axiom of choice). The cardinality of the underlying set of G_2 is |K|^{\alpha_2}, which is the power cardinality of \alpha_2 (assuming the axiom of choice). By assumption, these two power cardinals are distinct.