# (Q*,.) residually finite?

#### Banach

Hi! I have just shown that (Q,+) is not residually finite (Q the rational numbers). But what about (Q*,.) the group of rational units? I think it is residually finite:

Let x be a negative rational number, then x is not in the subgroup generated by all positive primes which has index 2. If x=a/b is a positive rational number such that the number of prime factors of a plus the number of prime factors of b is odd, then x is not in the subgroup generated by all negative primes which also has index 2. But i am stuck with the case where the number of prime factors of a plus the number of prime factors of b is even.

Does anybody know how to work that out?

Thankful for any help
Banach

#### Swlabr

Hi! I have just shown that (Q,+) is not residually finite (Q the rational numbers). But what about (Q*,.) the group of rational units? I think it is residually finite:

Let x be a negative rational number, then x is not in the subgroup generated by all positive primes which has index 2. If x=a/b is a positive rational number such that the number of prime factors of a plus the number of prime factors of b is odd, then x is not in the subgroup generated by all negative primes which also has index 2. But i am stuck with the case where the number of prime factors of a plus the number of prime factors of b is even.

Does anybody know how to work that out?

Thankful for any help
Banach
I do like the first result - it is actually much stronger than "not residually finite". What actually happens is that every element in $$\displaystyle \mathbb{Q}/H$$ has finite order while $$\displaystyle \mathbb{Q}/H$$ is infinite for every $$\displaystyle H \leq \mathbb{Q}$$ (this isn't what you would call easy to prove, but it is quite elementary).

For your second result, note that $$\displaystyle \frac{a}{b}\cdot \frac{c}{d} = \frac{ac}{bd}$$ and so you can get manipulate this to get a homomorphism...$$\displaystyle \frac{a}{b} \mapsto ab$$. Then make this into a finite homomorphism such that $$\displaystyle \frac{a}{b}$$ is not mapped to the identity.

#### Banach

How do you mean "manipulate this"? $$\displaystyle \frac{a}{b} \mapsto a \cdot b$$ is already a homomorphism from (Q*,.) to (Q*,.). And i also do not see yet how to transfer that to a finite group...

Can you give me another hint?

#### Swlabr

How do you mean "manipulate this"? $$\displaystyle \frac{a}{b} \mapsto a \cdot b$$ is already a homomorphism from (Q*,.) to (Q*,.). And i also do not see yet how to transfer that to a finite group...

Can you give me another hint?

What is the image of this homomorphism? It is a subgroup of $$\displaystyle \mathbb{Q}$$, yes, but it is not the whole of $$\displaystyle \mathbb{Q}$$.

(HINT: what are $$\displaystyle a$$ and $$\displaystyle b$$?)

#### Banach

Well, the image is $$\displaystyle \mathbb{Z}$$. I am thrilled where this is going. I think you want to consider something like U:= $$\displaystyle \{\frac{c}{d}:c \cdot d \equiv 1 (p)\}$$, where p is a prime that divides $$\displaystyle a \cdot b$$.

Or you want to factor out something? But (Z,.) is no group anymore...

#### tonio

Well, the image is $$\displaystyle \mathbb{Z}$$. I am thrilled where this is going. I think you want to consider something like U:= $$\displaystyle \{\frac{c}{d}:c \cdot d \equiv 1 (p)\}$$, where p is a prime that divides $$\displaystyle a \cdot b$$.

Or you want to factor out something? But (Z,.) is no group anymore...

I am thrilled myself...(Surprised)...I have a sketch of a proof but it's rather long and involved and still there's a little problem for even powers of 2...(Headbang)

Let us wait, perhaps Swlabr has a simpler proof...anyway, perhaps later I'll write down my piece.

Tonio

#### Swlabr

Or you want to factor out something? But (Z,.) is no group anymore...

A very good point, and one I seem to have missed...

What about quotienting out the subgroup generated by $$\displaystyle \frac{a+1}{b}$$. Clearly this will give you that $$\displaystyle \frac{a}{b}$$ is not equal to 1, and we then need to show that this group is finite...but I don't have time to think about this at the moment, sorry!

I'll try and reply again tomorrow.

#### Banach

@swlabr
This quotient won't be finite. But thank you anyway for helping me on that.

@tonio
maybe you could write down the proof you know. That would be very helpful.

But what about considering U. It is a subgroup and does not contain a/b. But does it have finite index? I am not so sure.

#### Swlabr

I am thrilled myself...(Surprised)...I have a sketch of a proof but it's rather long and involved and still there's a little problem for even powers of 2...(Headbang)

Let us wait, perhaps Swlabr has a simpler proof...anyway, perhaps later I'll write down my piece.

Tonio
Mine was a failed work in progress! However...a f.g. abelian group where every element has finite order is finite. So...

Let $$\displaystyle a/b$$ be the element we want to not be the identity. Quotient out the group generated by all the primes which do not divide $$\displaystyle a$$ from $$\displaystyle \mathbb{Q}$$. This will give us finitely generated.

To get elements of finite order...well, $$\displaystyle \frac{a}{b}, gcd(a, b)=1$$, is a power of some rational numbers, $$\displaystyle a=(\frac{p}{q})^n, gcd(p, q)=1$$. Take the largest such n, so for 4 n=2, but for 6 n=1. Then, quotient out the subgroup $$\displaystyle \{c^{2n}:c \in \mathbb{Q}\}$$.

That should work...although it was thought up on my way home so may not...

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#### Banach

Ok, I think I see why the quotient is finitely generated. But I do not get why every element has finite order!? Could you explain the second step please?

EDIT: Of course, now i see it. Thats why you quotient out the second group.

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