Every cyclic group is countable. Let \(\displaystyle G=\langle g\rangle\) be cyclic and define \(\displaystyle \theta:\mathbb{N}\to G:n\mapsto g^n\). Clearly \(\displaystyle \theta\) is surjective and so \(\displaystyle G\) is countable.

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.

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

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.

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.

I'm not sure what you mean by that. You will have to include 0 in \(\displaystyle \mathbb{N}\), as is often done. Then \(\displaystyle g^0= 1\) 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".

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 \(\displaystyle \mathbb{N}\), as is oftern done. Then \(\displaystyle g^0= 1\) 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".