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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.

Related facts

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.