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

, 01:11, 8 June 2012
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==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 $K_1</matH> with itself. The topology is the [[topospaces:product topology|product topology]] from the discrete topology of [itex]K_1$. Similarly, as a group, $G_2$ is the countable times unrestricted [[external direct product]] of $K_2</matH> with itself. The topology is the [[topospaces:product topology|product topology]] from the discrete topology of [itex]K_2$.
Then, we note that:
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