Is every isomorphism between groups preserves cardinality of elements?

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  • Thread starter Thread starter Mr Davis 97
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Discussion Overview

The discussion revolves around the properties of isomorphisms between groups, specifically whether the preservation of cardinality of elements is universally true for all elements in a group. Participants explore the implications of two statements regarding isomorphisms and cardinality, examining their equivalence and the conditions under which they hold.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant states that if ##\phi : G \to H## is an isomorphism, then ##|\phi(x)| = |x|## for all ##x \in G##, questioning if this is equivalent to the existence of some ##x \in G## for which the same holds.
  • Another participant asserts that both statements are always true and suggests they are equivalent, emphasizing that the requirement of isomorphism makes the distinction between "all" and "at least one" irrelevant.
  • A later reply expresses confusion about the equivalence of the two statements and mentions an attempt to reformulate the statements into prenex normal form, indicating a misunderstanding of the rules applied.
  • Another participant clarifies the implications of the two statements, noting that if the first statement is true for all ##x##, it must also be true for at least one ##x##, but raises conditions regarding the solution set for the cardinality equation.

Areas of Agreement / Disagreement

Participants express differing views on the equivalence of the two statements regarding isomorphisms and cardinality. While some argue they are equivalent, others highlight the nuances that may lead to differing interpretations, indicating that the discussion remains unresolved.

Contextual Notes

Participants reference the need for clarity in the implications of quantifiers in logical statements, suggesting that misunderstandings may arise from the application of rules related to prenex normal form. There is also mention of conditions that affect the truth of the statements, which remain unspecified.

Mr Davis 97
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I have the following statement: If ##\phi : G \to H## is an isomorphism, then ##|\phi(x) | = |x|## for all ##x \in G##. Is this equivalent to the following? There exists an ##x \in G## such that if ##\phi : G \to H## is an isomorphism, then ##|\phi(x) | = |x|##.
 
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Mr Davis 97 said:
I have the following statement: If ##\phi : G \to H## is an isomorphism, then ##|\phi(x) | = |x|## for all ##x \in G##. Is this equivalent to the following? There exists an ##x \in G## such that if ##\phi : G \to H## is an isomorphism, then ##|\phi(x) | = |x|##.
They are both always true, so in a way they are equivalent. What do you really want to know? As soon as you have the requirement "isomorphism" it doesn't matter anymore whether there is just one ##x##, because there are always all.
 
fresh_42 said:
They are both always true, so in a way they are equivalent. What do you really want to know? As soon as you have the requirement "isomorphism" it doesn't matter anymore whether there is just one ##x##, because there are always all.
I'm just confused about how the two statement are saying the same thing. Basically what I am trying to do is to put the statement into prenex normal form and then to interpret how the statements are equivalent. I'm using the rule for taking the quantifier of the consequent that is described here: https://en.wikipedia.org/wiki/Prenex_normal_form#Implication

Edit: Actually, I used the wrong rule to begin with, so in this case my question is meaningless. If I come across another example I'll ask.
 
The first statement says that for all x (it is true that if phi is an isomorphism then abs(phi(x))=abs(x)). The second statement says that for at least one x (it is true that if phi is an isomorphism then abs(phi(x))=abs(x)). If the first statement is true for all x then certainly it is true for at least one x (with the condition that the set of solutions for x in the formula abs(phi(x))=abs(x) must be nonempty.In the converse case the first statement only follows from second if the size of the solution set for abs(phi(x))=abs(x) is less than equal or less than one.
 

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