Undergrad Is every isomorphism between groups preserves cardinality of elements?

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The discussion centers on the equivalence of two statements regarding isomorphisms between groups and the preservation of cardinality of elements. The first statement asserts that for all elements x in group G, an isomorphism φ ensures that the cardinality of φ(x) equals the cardinality of x. The second statement claims that there exists at least one element x in G for which the same holds true under an isomorphism. It is clarified that if the first statement is true for all x, it must also be true for at least one x, making them equivalent under certain conditions. The confusion arose from the initial interpretation of the statements, which was resolved by recognizing the implications of quantifiers in logical expressions.
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.
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

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