# subgroup question

• Sep 18th 2010, 07:06 PM
Danneedshelp
subgroup question
Q: Find all possible finite subgroups of the non negative rational numbers under multiplication.

I can only think of the singleton set containing the identity element which would be of order 1. All other sets would be of infinite order. I feel there is more to this though.

thanks
• Sep 18th 2010, 07:28 PM
tonio
Quote:

Originally Posted by Danneedshelp
Q: Find all possible finite subgroups of the non negative rational numbers under multiplication.

I can only think of the singleton set containing the identity element which would be of order 1. All other sets would be of infinite order. I feel there is more to this though.

thanks

The non-negative rationals under multiplication are not a group...the POSITIVE rationals are, though.

As for your solution it is correct: if $\displaystyle 0<q\in\mathbb{Q}\,,\,\,then\,\,\{q^n\}_{n=1}^\inft y$ is an infinite set...

Tonio
• Sep 18th 2010, 07:35 PM
undefined
Quote:

Originally Posted by tonio
The non-negative rationals under multiplication are not a group...the POSITIVE rationals are, though.

As for your solution it is correct: if $\displaystyle 0<q\in\mathbb{Q}\,,\,\,then\,\,\{q^n\}_{n=1}^\inft y$ is an infinite set...

Tonio

Is it supposed to be $\displaystyle 1\ne q\in\mathbb{Q}$ instead of $\displaystyle 0<q\in\mathbb{Q}$?

Edit: Meh I was thinking of $\displaystyle \,\mathbb{Q}$ as the positive rationals, should have rewritten the set or instead restricted $\displaystyle q\not\in\{-1,0,1\}$.
• Sep 18th 2010, 07:38 PM
Danneedshelp
Quote:

Originally Posted by tonio
The non-negative rationals under multiplication are not a group...the POSITIVE rationals are, though.

As for your solution it is correct: if $\displaystyle 0<q\in\mathbb{Q}\,,\,\,then\,\,\{q^n\}_{n=1}^\inft y$ is an infinite set...

Tonio

Oh, my bad. I don't know why I wrote that.

Should it be for all $\displaystyle q\in{\mathbb{Q}}-\{0\}$ and $\displaystyle q\neq\\e$?

Thanks for the help
• Sep 19th 2010, 04:41 AM
tonio
Quote:

Originally Posted by Danneedshelp
Oh, my bad. I don't know why I wrote that.

Should it be for all $\displaystyle q\in{\mathbb{Q}}-\{0\}$ and $\displaystyle q\neq\\e$?

Thanks for the help

Yes, but it is also true for the positive rationals only (without 1)