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

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

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

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.