# how to prove this statement in modern algebra (finite abelian group)

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.

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In your first statement it must be $\,\Bbb Q^*:=\Bbb Q-\{0\}\,$ – DonAntonio Nov 14 '12 at 3:01
@DonAntonio thats equivalent – math Nov 14 '12 at 3:08
Hint: show that $\,\Bbb Q^*_+\cong \bigoplus_{p\,\text{ a prime}}\langle\,p\,\rangle\,$ – DonAntonio Nov 14 '12 at 3:11
@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 ..." – DonAntonio Nov 14 '12 at 3:12
@DonAntonio could u please eloberate this u will be appreciated – math Nov 14 '12 at 3:12

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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i have one concern about this proof. How could we say that the basis is linearly independent. – math Nov 15 '12 at 2:12
@madhavphuyal Do you know the definition of linear independence of the basis? – Makoto Kato Nov 15 '12 at 2:18
because 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. – math Nov 15 '12 at 4:40
@motu What exactly you don't understand?. – Makoto Kato Nov 15 '12 at 4:48
how do u plug in the last statement. because we have basis are not of the form. – math Nov 15 '12 at 4:52

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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is this the complete proof or its just the outline – math Nov 14 '12 at 4:23
@madhavphuyal It is not the outline. Please feel free to ask me about the proof. – Makoto Kato Nov 14 '12 at 4:26
how u justified the universal properties of free abelian group in this problem – math Nov 14 '12 at 4:40
@madhavphuyal I did not use the universal properties of free abelian group. I just proved that $G$ is isomorphic to $F$. – Makoto Kato Nov 14 '12 at 5:05
@ Makoto Kato Thank u very much – math Nov 14 '12 at 5:11