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

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.
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.