# Difference between revisions of "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.

## 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 $3^\infty$.