subgroup question

**Danneedshelp** · September 18th 2010, 07:06 PM

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

**tonio** · September 18th 2010, 07:28 PM

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 $0<q\in\mathbb{Q}\,,\,\,then\,\,\{q^n\}_{n=1}^\inft y$ is an infinite set...

Tonio

**undefined** · September 18th 2010, 07:35 PM

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 $0<q\in\mathbb{Q}\,,\,\,then\,\,\{q^n\}_{n=1}^\inft y$ is an infinite set...

Tonio

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

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

**Danneedshelp** · September 18th 2010, 07:38 PM

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 $0<q\in\mathbb{Q}\,,\,\,then\,\,\{q^n\}_{n=1}^\inft y$ is an infinite set...

Tonio

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

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

Thanks for the help

**tonio** · September 19th 2010, 04:41 AM

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)

Thread: subgroup question

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