I figure out that Q set of rational number under multiplication form a group but i am stuck in this part could some one show me way out. Q* the set of all positive rational number forms a free abelian group binary operation under multiplication.
how to prove this statement in modern algebra (finite abelian group)
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0In your first statement it must be $\,\Bbb Q^*:=\Bbb Q-\{0\}\,$ – 2012-11-14
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0@DonAntonio thats equivalent – 2012-11-14
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0Hint: show that $\,\Bbb Q^*_+\cong \bigoplus_{p\,\text{ a prime}}\langle\,p\,\rangle\,$ – 2012-11-14
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0@Madhave, you missed my point: you wrote "the set of rational numbers under multiplication...", and it must be "the set of *non-zero* rational numbers ..." – 2012-11-14
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0@DonAntonio could u please eloberate this u will be appreciated – 2012-11-14
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0@DonAntonio it is the statement i copied exactly from Thomos Hungerford book – 2012-11-14
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0a field is NOT a multiplicative group since the element zero has not multiplicative inverse. Every field *without the zero* element is a multiplicative group. – 2012-11-14
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0Give me please the number of exercise, chapter, page ,etc in Hungerford's book, @madhav. – 2012-11-14
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0@DonAntonio pg 75 question 11 (b) – 2012-11-14
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0Chapter 2 free abelian group – 2012-11-14
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0Ok Madhav, exactly what I said: Hungerford writes $\,\Bbb Q^*\,$ which means the *non-zero* rationals. – 2012-11-14
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0Anyway, read my second comment here for a nice hint. – 2012-11-14
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0@DonAntonio yes i have also mention the Q* in my statement any way could you helo me to solve this – 2012-11-14
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0Which part are you having trouble with? Proving it's a group? Proving it's abelian? Proving it's free? – 2012-11-14
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0@GerryMyerson Free abelian group. i am not sure with the solution below so i need some clerification on free abelian group – 2012-11-14
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0So, you can prove it's an abelain group, but you can't prove it's free? or you can prove it's a free group, but you can't prove it's abelian? or you can prove it's free abelian, but you can't prove it's a group? Come on, now --- which one or ones of the three concepts is it that's giving you trouble? – 2012-11-14
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0@GerryMyerson it should be a free abelian group thats my need – 2012-11-14
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0OK, let's try this again. Can you prove that it's abelian? – 2012-11-14
2 Answers
At the request of the OP, I will show a direct proof. Let $G$ be the set of positive rational numbers. $G$ is a group under mutiplications. Let $\Pi$ be the set of prime numbers. We claim $G$ is a free abelian group with a basis $\Pi$.
Let $r \in G$. there exist positive integers $a, b$ such that $r = \frac{a}{b}$. Since $a$ and $b$ can be written as products of prime numbers, $G$ is generated by $\Pi$.
Let $p_1,\dots,p_r$ be distict prime numbers. Suppose $p_1^{n_1}\cdots p_r^{n_r} = 1$, where all $n_i$ are integers. It suffices to prove that all $n_i = 0$. If all $n_i \ge 0$, clearly all $n_i = 0$. Hence we assume not all $n_i \ge 0$. Without loss of generality, we can assume that $n_1, \dots, n_k \ge 0$ and $n_{k+1},\dots n_r < 0$. Then $p_1^{n_1}\cdots p_k^{n_k} = p_{k+1}^{-n_{k+1}}\cdots p_r^{-n_r}$. But this is a contradiction because $\mathbb{Z}$ is a UFD.
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0i have one concern about this proof. How could we say that the basis is linearly independent. – 2012-11-15
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0@madhavphuyal Do you know the definition of linear independence of the basis? – 2012-11-15
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0because i am thinking about the result p n 1 1 ⋯p n k k =p −n k+1 k+1 ⋯p −n r. isn't it need some clearification. – 2012-11-15
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0@motu What exactly you don't understand?. – 2012-11-15
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0how do u plug in the last statement. because we have basis are not of the form. – 2012-11-15
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0@motu Do you know that every integer greater than 1 can be uniquely(except orders) written as a product of prime numbers? – 2012-11-15
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0Ohhhhhh........thats i missed. thanks buddy. I appreciate ur help. good night see u again – 2012-11-15
Let $G$ be the set of positive rational numbers. $G$ is a group under mutiplications. Let $\Pi$ be the set of prime numbers. We claim $G$ is a free abelian group with a basis $\Pi$. Let $F$ be the free abelian group generated by $\Pi$. Every $x \in F$ can be written uniquely as $x = \sum_i n_ip_i$, where $p_i'$s are distinct prime numbers. Let $f\colon F \rightarrow G$ be the map defined by $f(x) = \prod_i p_i^{n_i}$. Clearly $f$ is a homomorphism. Since $\mathbb{Z}$ is a UFD, $f$ is injective.
It remains to prove that $f$ is surjective. Let $r \in G$. there exist positive integers $a, b$ such that $r = \frac{a}{b}$ and gcd$(a, b) = 1$. $a$ can be uniquely written $a = (p_1)^{n_1}\cdots (p_r)^{n_r}$, where $p_i'$s are distinct prime numbers. Similarly $b$ can be uniquely written $b = (q_1)^{m_1}\cdots (q_s)^{m_s}$, where $q_i'$s are distinct prime numbers. Since gcd$(a, b) = 1$, $p_i'$s and $q_j'$s are distinct. Let $x = \sum_i n_ip_i - \sum m_jq_j$. Clearly $f(x) = r$. Hence $f$ is surjective.
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0is this the complete proof or its just the outline – 2012-11-14
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0@madhavphuyal It is not the outline. Please feel free to ask me about the proof. – 2012-11-14
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0how u justified the universal properties of free abelian group in this problem – 2012-11-14
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0@madhavphuyal I did not use the universal properties of free abelian group. I just proved that $G$ is isomorphic to $F$. – 2012-11-14
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0@ Makoto Kato Thank u very much – 2012-11-14
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0we asked to prove this is free abelian group but we start from that point is my concern. is there another way of proving this questin – 2012-11-14
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0@madhavphuyal One can prove that $G$ is a free abelian group more straightly, but I think it would be essentially the same as the one above. If you want me to prove it more straightly, I will do so. – 2012-11-14
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0that could be better thak u very much – 2012-11-14