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 $$G_1$$ 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

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

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 $$3^\infty$$.