So if a subgroup is not countable, then it is not cyclic.
so in this case, if the element in the group has infinite order, then it is not cyclic..but im not sure how does that help in writing a proof for this question
in the function that you have defined, wont there be no identity under multiplication? im not sure how to define functions and always wonder how are they defined.
No, infinite is not enough- it must be uncountable. The set of positive real numbers is uncountable, therefore any group having the set of positive real number as underlying set, in particular the "set of positive real numbers under multiplication" is not cyclic.
I'm not sure what you mean by that. You will have to include 0 in , as is often done. Then is the identity. Note that this is simply a "one-to-one" function from the non-negative integers to the group. It is not claimed to be an "isomorphism".in the function that you have defined, wont there be no identity under multiplication? im not sure how to define functions and always wonder how are they defined.
Usually the uncountability of the real numbers is a standard fact that you can use. However, if you need to prove it, see Wikipedia:
Cantor's diagonal argument - Wikipedia, the free encyclopedia