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

## Statement

It is possible to have two profinite groups and that have the same order as each other as abstract groups but such that the order as a profinite group for is not the same as the order as a profinite group for . Note that the equality of orders as profinite groups is checked in the sense of equality of supernatural numbers.

## Proof

Let be cyclic group:Z2 and be cyclic group:Z3. Consider the groups and . As a group, is the countable times unrestricted external direct product of with itself. The topology is the product topology from the discrete topology of . Similarly, as a group, is the countable times unrestricted external direct product of with itself. The topology is the product topology from the discrete topology of .

Then, we note that:

- The cardinalities of the underlying sets of and are equal: Th cardinality of the underlying set of is , which is the cardinality of the continuum, i.e., the power cardinal of the first infinite ordinal. The cardinality of the underlying set of is , 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 is not equal, as a supernatural number, to the order (as a profinite gorup) of : The order of is and the order of is .