I Is every isomorphism between groups preserves cardinality of elements?

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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.
 
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