Cardinality of underlying set of a profinite group need not determine order as a profinite group

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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 as abstract groups but such that the order as a profinite group for G_1 is not the same as the order as a profinite group for G_2. Note that the equality of orders as profinite groups is checked in the sense of equality of supernatural numbers.

Related facts

Proof

Let K_1 be cyclic group:Z2 and K_2 be cyclic group:Z3. Consider the groups G_1 = K_1^\omega and G_2 = K_2^\omega. As a group, G_1 is the countable times unrestricted external direct product of K_1 with itself. The topology is the product topology from the discrete topology of K_1. Similarly, as a group, G_2 is the countable times unrestricted external direct product of K_2 with itself. The topology is the product topology from the discrete topology of K_2.

Then, we note that:

  • The cardinalities of the underlying sets of G_1 and G_2 are equal: Th cardinality of the underlying set of G_1 is 2^\omega, which is the cardinality of the continuum, i.e., the power cardinal of the first infinite ordinal. The cardinality of the underlying set of G_2 is 3^\omega = 2^\omega, which is the cardinality of the continuum, i.e., the power cardinal of the first infinite ordinal. Thus both cardinals are equals.
  • The order (as a profinite group) of G_1 is not equal, as a supernatural number, to the order (as a profinite gorup) of G_2: The order of G_1 is 2^\infty and the order of G_2 is 3^\infty.