cyclic

Aug 2009
639
2
how do you prove that the set of positive real numbers under multiplication is not cyclic

in general, how do you prove that something is not cyclic?

thanks
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
how do you prove that the set of positive real numbers under multiplication is not cyclic

in general, how do you prove that something is not cyclic?

thanks
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, why's that help us?
 
Aug 2009
639
2
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.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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".
 
Last edited:
Aug 2009
639
2
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".
so to form a proof for it, i have to show that it is uncountable? how do i do that?