# cyclic

• May 28th 2010, 02:44 PM
alexandrabel90
cyclic
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
• May 28th 2010, 03:02 PM
Drexel28
Quote:

Originally Posted by alexandrabel90
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?
• May 29th 2010, 12:41 AM
alexandrabel90
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.
• May 29th 2010, 02:07 AM
HallsofIvy
Quote:

Originally Posted by alexandrabel90
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.

Quote:

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".
• May 29th 2010, 03:43 AM
alexandrabel90
Quote:

Originally Posted by HallsofIvy
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?
• May 29th 2010, 06:28 AM
roninpro
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