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Math Help - (Q*,.) residually finite?

  1. #1
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    (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
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Banach View Post
    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 \mathbb{Q}/H has finite order while \mathbb{Q}/H is infinite for every 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 \frac{a}{b}\cdot \frac{c}{d} = \frac{ac}{bd} and so you can get manipulate this to get a homomorphism... \frac{a}{b} \mapsto ab. Then make this into a finite homomorphism such that \frac{a}{b} is not mapped to the identity.
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    Thanks for your answer.
    How do you mean "manipulate this"? \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?




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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Banach View Post
    Thanks for your answer.
    How do you mean "manipulate this"? \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 \mathbb{Q}, yes, but it is not the whole of \mathbb{Q}.

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

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

    Help me, please
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    Quote Originally Posted by Banach View Post
    Well, the image is \mathbb{Z}. I am thrilled where this is going. I think you want to consider something like U:= \{\frac{c}{d}:c \cdot d \equiv 1 (p)\}, where p is a prime that divides a \cdot b.

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

    Help me, please

    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
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    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Banach View Post
    Or you want to factor out something? But (Z,.) is no group anymore...

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

    What about quotienting out the subgroup generated by \frac{a+1}{b}. Clearly this will give you that \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.
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    @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.
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    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by tonio View Post
    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 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 a from \mathbb{Q}. This will give us finitely generated.

    To get elements of finite order...well, \frac{a}{b}, gcd(a, b)=1, is a power of some rational numbers, 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 \{c^{2n}:c \in \mathbb{Q}\}.

    That should work...although it was thought up on my way home so may not...
    Last edited by Swlabr; May 11th 2010 at 01:46 AM. Reason: Small logical error
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    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.
    Last edited by Banach; May 10th 2010 at 10:50 AM.
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    let H=\{1,-1\} \lhd \mathbb{Q}^{\times} and suppose p_j is the j-th prime number. define the map f: \mathbb{Q}^{\times} \to H \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus \cdots by f(r)=(\text{sgn}(r), n_1,n_2, \cdots , n_k, 0, 0, \cdots), where r=\pm p_1^{n_1} p_2^{n_2} \cdots p_k^{n_k}.

    clearly 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!)
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    Thumbs up

    Thats really elegant.

    Thank you very much!
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    Quote Originally Posted by Banach View Post
    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 p_i-adic evaluation of every non-zero rational, which is the way the paper proposes to show the existence of maximal subgroups of \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 \mathbb{Q}^{*} that doesn't contain 4... ...of course, I haven't yet read carefully the paper.

    Tonio
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    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Banach View Post
    Well, the image is \mathbb{Z}. I am thrilled where this is going. I think you want to consider something like U:= \{\frac{c}{d}:c \cdot d \equiv 1 (p)\}, where p is a prime that divides a \cdot b.

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

    Help me, please
    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 (\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 \mathbb{Q} is residually finite, then as it's Schutzenberger groups are just \{0\} and \mathbb{Q} \setminus \{0\} we have that \mathbb{Q}^{\ast} is a residually finite group.

    I think that is actually quite a nice result...
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    Quote Originally Posted by tonio View Post
    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 \mathbb{Q}^{*} that doesn't contain 4... ...of course, I haven't yet read carefully the paper.
    the map f that i defined gives a subgroup with that property quite fast and easily: for example f(4)=(1,2,0,0, \cdots). now let K=H \oplus 3\mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} \cdots . then N=f^{-1}(K) would be a

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