Difference between revisions of "Cardinality of underlying set of a profinite group need not determine order as a profinite group"

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(Created page with "==Statement== It is possible to have two profinite groups <math>G_1</math> and <math>G_2</math> that have the same [[order of a group|order as each other as abstract grou...")
 
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==Proof==
 
==Proof==
  
Let <math>K_1</math> be [[cyclic group:Z2]] and <math>K_2</math> be [[cyclic group:Z3]]. Consider the groups <math>G_1 = K_1^\omega</math> and <math>G_2 = K_2^\omega</math>. As a group, <math>G_1</math> is the countable times unrestricted [[external direct product]] of <math>K_1</matH> with itself. The topology is the [[topospaces:product topology]] from the discrete topology of <math>K_1</math>.  Similarly, as a group, <math>G_2</math> is the countable times unrestricted [[external direct product]] of <math>K_2</matH> with itself. The topology is the [[topospaces:product topology]] from the discrete topology of <math>K_2</math>.
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Let <math>K_1</math> be [[cyclic group:Z2]] and <math>K_2</math> be [[cyclic group:Z3]]. Consider the groups <math>G_1 = K_1^\omega</math> and <math>G_2 = K_2^\omega</math>. As a group, <math>G_1</math> is the countable times unrestricted [[external direct product]] of <math>K_1</matH> with itself. The topology is the [[topospaces:product topology|product topology]] from the discrete topology of <math>K_1</math>.  Similarly, as a group, <math>G_2</math> is the countable times unrestricted [[external direct product]] of <math>K_2</matH> with itself. The topology is the [[topospaces:product topology|product topology]] from the discrete topology of <math>K_2</math>.
  
 
Then, we note that:
 
Then, we note that:

Revision as of 01:11, 8 June 2012

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.