1. ## 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

2. 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 $0 is an infinite set...

Tonio

3. 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 $0 is an infinite set...

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

Edit: Meh I was thinking of $\,\mathbb{Q}$ as the positive rationals, should have rewritten the set or instead restricted $q\not\in\{-1,0,1\}$.

4. 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 $0 is an infinite set...

Tonio

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

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

Thanks for the help

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

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

Thanks for the help

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