free groups, finitely generated groups

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May 23rd 2009, 12:49 AM
georgel

free groups, finitely generated groups

I have three problems I wasn't able to solve by myself. :(

1) Is ( $\mathbb{R}^{*}, .$ ) finitely generated group? Is it free?

2) Is $\mathbb{Z}_3 \oplus \mathbb{Z}$ free Abelian group?

3) Is $\mathbb{Z}_6$ free group?

I know that in order to check whether a group is finitely generated, I have to find a finite set that generates it, or prove it doesn't exist, but I can't do either. :(

Also, to check whether something is a free group, I would have to find its basis.
For example, is $\{0, 1, 2, 3, 4, 5 \}$ the basis of $\mathbb{Z}_6$ ?

I'm really lost, please help.
May 23rd 2009, 01:19 AM
NonCommAlg

Quote:

Originally Posted by georgel

I have three problems I wasn't able to solve by myself. :(

1) Is ( $\mathbb{R}^{*}, .$ ) finitely generated group? Is it free?

it's not free because it's not torsion-free. for example $-1$ has finite order. (recall that every free abelian group is torsion-free)

it's not finitely generated because $\mathbb{Q}^*,$ which is a subgroup of $\mathbb{R}^*,$ is not finitely generated. (recall that every subgroup of a finitely generated abelian group is finitely generated)

Quote:

2) Is $\mathbb{Z}_3 \oplus \mathbb{Z}$ free Abelian group?

it's not free because it's not torsion-free. for example $(1,0)$ has finite order.

it is finitely generated because it's generated by $(1,0)$ and $(0,1).$

Quote:

3) Is $\mathbb{Z}_6$ free group?

no it's not! a non-trivial finite group is never free!
May 23rd 2009, 02:01 AM
georgel

Thank you so much!!!

Quote:

Originally Posted by NonCommAlg

no it's not! a non-trivial finite group is never free!

Could you possibly explain why this is so?
May 23rd 2009, 02:04 AM
Swlabr

Quote:

Originally Posted by georgel

1) Is ( $\mathbb{R}^{*}, .$ ) finitely generated group? Is it free?

Further to NCA's answer, I would like to point out that no uncountably infinite group is finitely generated. This is because every product is, by definition, of finite length and so all products can be "counted" using the natural numbers.

$\mathbb{R}$ is uncountably infinite and so is not finitely generated.
May 23rd 2009, 02:35 AM
NonCommAlg

Quote:

Originally Posted by georgel

Thank you so much!!!

Could you possibly explain why this is so?

in a free group, the elements of a basis they all have infinite order but a finite group has no element of infinite order!

what Swlabr said is true and it works for your problem, but it doesn't prove that $\mathbb{Q}^*$ is not finitely generated (which is a good exercise! (Nod)).
May 23rd 2009, 03:31 AM
georgel

Quote:

Originally Posted by NonCommAlg

in a free group, the elements of a basis they all have infinite order but a finite group has no element of infinite order!

what Swlabr said is true and it works for your problem, but it doesn't prove that $\mathbb{Q}^*$ is not finitely generated (which is a good exercise! (Nod)).

I'm terribly ashamed I din't think of that. :(

Thanks to both you and Swlabr!

And I already proved that $\mathbb{Q}^*$ is not finitely generated! :)