1. ## (Q*,.) residually finite?

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

2. Originally Posted by 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
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.

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?

4. Originally Posted by 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?

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 $\displaystyle b$?)

5. 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...

6. Originally Posted by 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...

I am thrilled myself......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...

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

Tonio

7. Originally Posted by Banach
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.

8. @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.

9. Originally Posted by tonio
I am thrilled myself......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...

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...

10. 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.

11. let $\displaystyle H=\{1,-1\} \lhd \mathbb{Q}^{\times}$ and suppose $\displaystyle p_j$ is the j-th prime number. define the map $\displaystyle f: \mathbb{Q}^{\times} \to H \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus \cdots$ by $\displaystyle f(r)=(\text{sgn}(r), n_1,n_2, \cdots , n_k, 0, 0, \cdots),$ where $\displaystyle r=\pm p_1^{n_1} p_2^{n_2} \cdots p_k^{n_k}.$

clearly $\displaystyle f$ is an isomorphism and now the result follows because the direct product of a family of residually finite groups is residually finite. (very easy to prove!)

12. Thats really elegant.

Thank you very much!

13. Originally Posted by Banach
Thats really elegant.

Thank you very much!

NCA's solution is practically the same I was thinking of based on the paper http://www.mathematik.uni-bielefeld.de/lag/man/061.pdf ( look lemma 1 at the end of page 2 )

NCA basically showed the map defined by the $\displaystyle p_i-$adic evaluation of every non-zero rational, which is the way the paper proposes to show the existence of maximal subgroups of $\displaystyle \mathbb{Q}^{*}$ ...I still have the little doubt about positive rationals which are even powers of 2 but that's due, perhaps , to my lack of understanding: I haven't yet found a finite-index subgroup of $\displaystyle \mathbb{Q}^{*}$ that doesn't contain 4... ...of course, I haven't yet read carefully the paper.

Tonio

14. Originally Posted by 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...

I have done a bit of reading, and I believe this technique actually works.

So, what I was trying to do was prove that the monoid $\displaystyle (\mathbb{Q}, .)$ is residually finite.

Then...there is a theorem which says that the groups of H-classes (Schutzenberger groups) of a monoid M are residually finite if and only if the monoid itself is residually finite. So...if we have proved that the monoid $\displaystyle \mathbb{Q}$ is residually finite, then as it's Schutzenberger groups are just$\displaystyle \{0\}$ and $\displaystyle \mathbb{Q} \setminus \{0\}$ we have that $\displaystyle \mathbb{Q}^{\ast}$ is a residually finite group.

I think that is actually quite a nice result...

15. Originally Posted by tonio
I still have the little doubt about positive rationals which are even powers of 2 but that's due, perhaps , to my lack of understanding: I haven't yet found a finite-index subgroup of $\displaystyle \mathbb{Q}^{*}$ that doesn't contain 4... ...of course, I haven't yet read carefully the paper.
the map $\displaystyle f$ that i defined gives a subgroup with that property quite fast and easily: for example $\displaystyle f(4)=(1,2,0,0, \cdots).$ now let $\displaystyle K=H \oplus 3\mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} \cdots .$ then $\displaystyle N=f^{-1}(K)$ would be a

subgroup that you're looking for. in fact $\displaystyle N$ is the set of all elements of $\displaystyle r \in \mathbb{Q}^{\times}$ such that $\displaystyle r=\pm 2^{3n_1}3^{n_2} \cdots p_s^{n_s},$ for some integers $\displaystyle n_j, \ s > 0.$ clearly $\displaystyle 4 \notin N$ and $\displaystyle [\mathbb{Q}^{\times}:N]=3.$